Fermat's Last Theorem and Related Problems
Created | Updated Dec 28, 2008
The mathematical background needed to read this article is usually elementary. A few prerequisites are: A natural number is a positive integer. A non-zero integer x is said to divide an integer y if y is an integer multiple of x. An integer x is said to be congruent to an integer y modulo a non-zero integer z (denoted by x≡y(mod z)) if z divides x-y. Integers x and y are said to be relatively prime if they have no common integer divisors other than 1 or -1. x+y, x+y≠0, where x and y are integers divides xn+yn where n is an odd natural number. If an odd prime p divides x+y where x and y are integers, then p2 divides xp+yp. A prime p divides xp-x where x is an integer (Fermat's "little" theorem). If x and y are relatively prime integers, then xφ(y)≡1(mod y) where φ(y) is Euler's "phi" function. The greatest common divisor of natural numbers x and y will be denoted by g.c.d.(x, y). A more succinct way of saying that an integer x is congruent to the pth power of an integer modulo an integer y is to say that x is a pth power modulo y.
(1) If p>3, there do not exist a and b such that [(ap+bp)/(a+b)] is a pth power.
(2) If p>3, there do not exist a and b such that [(ap+bp)/(a+b)] is a pth power, 2p does not divide a, b, a-b, or a+b, and a, b, a-b, or [a+b] is a pth power.
Note that if the first conjecture is true, there are no solutions of Fermat's equation ap+bp=cp (which, of course, is already known), and that the second conjecture encompasses the first case of Fermat's Last Theorem (where p does not divide abc). (In 1810, Barlow proved that ap+bp=cp only if [(ap+bp)/(a+b)] is a pth power.) Let T be a natural number. Since a, b, and T are not symmetrical in the equation [(ap+bp)/(a+b)]=Tp, it is not obvious how to apply the theory of elliptic curves to these problems. The "pth power w.r.t." concept and the identity ap+bp=(ap-bp)+2bp play a central role in proving these conjectures. For example, if p divides a+b and (ap+bp)/(a+b)/p is a pth power, then p(a+b)/2 is a pth power w.r.t. (ap-bp)/(a-b) (this is a succinct way of saying that (p(a+b))(f-1)/p≡2(f-1)/p(mod f) for every prime factor f, f≠p, of (ap-bp)/(a-b)). Note that if p is not a pth power modulo a prime f of the form pk+1, then, for example, if f does not divide a+b, (px(a+b))(f-1)/p≡1(mod f), 0≤x<p, has a solution (x defines a congruence class). Furthermore, if 2 is not a pth power modulo f, then, for example, (2yp2(a+b))(f-1)/p≡1(mod f), 0≤y<p, has a solution and a+b can be eliminated from the congruences. The objective in the following is to eliminate a, b, a-b, and a+b from certain congruences so that congruence relationships involving only 2 and p are obtained. When p=3, there do exist a and b such that [(ap+bp)/(a+b)] is a pth power and many properties of such a and b, among them reformulated versions of the classical Furtwängler and Vandiver theorems for Fermat's equation, can be empirically derived. In the following, these "propositions" are stated as if they were true for all p. One justification for doing this is the first conjecture above. Another justification for doing this for some of the propositions is the "difficult proof" hypothesis (not rigorous, of course). This hypothesis is; a, b, and T satisfying the equation [(ap+bp)/(a+b)]=Tp and having characteristic properties of the equation ap+bp=cp will have other properties in common with a, b, and c satisfying the equation ap+bp=cp (otherwise, there would be a contradiction and there would be a "simple" proof of Fermat's Last Theorem). For example, it can be easily proved that ap+bp=cp implies 2 is a pth power w.r.t. (ap-bp)/(a-b). Based on empirical evidence collected for p=3, if [(ap+bp)/(a+b)] is a pth power and 2p divides a+b, then 2 is a pth power w.r.t. (ap-bp)/(a-b) (although there is no apparent reason why this should be true [other than the "difficult proof" hypothesis]). More will be said on what these characteristic properties are later.
(3) If [(ap+bp)/(a+b)] is a pth power and 2p does not divide a, b, a-b, or a+b, then 2 and p are not pth powers w.r.t. (ap-bp)/(a-b).
(4) If [(ap+bp)/(a+b)] is a pth power, 2p does not divide a, b, a-b, or a+b, and f is a prime factor of [(ap-bp)/(a-b)], then 2 is a pth power modulo f if and only if f is of the form p2k+1.
By these two propositions, if [(ap+bp)/(a+b)] is a pth power and 2p does not divide a, b, a-b, or a+b, then there is at least one prime factor of [(ap-bp)/(a-b)] not of the form p2k+1. In 1912, Furtwängler proved that if ap+bp=cp, q divides a and p does not divide ac, or q divides b and p does not divide bc, then qp-1≡1(mod p2). (Proofs of this theorem use the condition that a+b must be a pth power and it is not obvious how to prove a reformulated version of the theorem without using this condition.) Furtwängler also proved that if ap+bp=cp, q divides a-b or a+b, and p does not divide a-b or a+b, then qp-1≡1(mod p2). Note that if p does not divide a natural number d, then dp(p-1)≡1(mod p2) by Euler's theorem. Then if qp-1≡1(mod p2), q is a pth power modulo p2. The reformulated version of Furtwängler's theorems is;
(5) If [(ap+bp)/(a+b)] is a pth power and 2 does not divide a, then p does not divide a and every prime factor of a is a pth power modulo p2. If [(ap+bp)/(a+b)] is a pth power, 2 divides a, and p does not divide a, then a/2 is a pth power modulo p2. (Analogous results hold for b.) If [(ap+bp)/(a+b)] is a pth power and 2 does not divide a-b, then p does not divide a-b and every prime factor of a-b is a pth power modulo p2. If [(ap+bp)/(a+b)] is a pth power and 2 does not divide a+b, then p2 does not divide a+b and every prime factor of a+b other than p (if p divides a+b) is a pth power modulo p2.
The peculiar form of Furtwängler's second theorem, that is, the condition that p not divide a-b or a+b, makes sense when viewed from this perspective; 2 divides a-b if and only if 2 divides a+b. This proposition implies that if [(ap+bp)/(a+b)] is a pth power, p divides a, b, a-b, or a+b, and 2p does not divide a, b, a-b, or a+b, then 2 divides a or b, and p divides a+b. Note that this proposition implies "split" 2 and p are not possible when ap+bp=cp, p divides c, since p2 cannot divide a+b in this case. (By Barlow's formulas, p(a+b) must be a pth power when ap+bp=cp, p divides c.) The requirement that 2p divide a+b could be said to be a characteristic property of the equation ap+bp=cp, p divides c. A related proposition is;
(6) If [(ap+bp)/(a+b)] is a pth power, then p2 divides a if 2p divides a, p2 divides b if 2p divides b, p2 divides a-b if 2p divides a-b, or p3 divides a+b if 2p divides a+b.
In the following, p/2 will be said to be a pth power modulo a prime f if p(f-1)/p≡2(f-1)/p(mod f). More propositions are;
(7) If [(ap+bp)/(a+b)] is a pth power, 2 divides a or b, and f is a prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, then exactly one of 2p, p, or p/2 is a pth power modulo f.
(8) If [(ap+bp)/(a+b)] is a pth power, p divides a+b, and f is a prime factor of [(ap-bp)/(a-b)] of the form p2k+1, then pa, pb, p2(a-b), and p(a+b) are pth powers modulo f. If [(ap+bp)/(a+b)] is a pth power, p divides a, b, or a-b, and f is a prime factor of [(ap-bp)/(a-b)] of the form p2k+1, then a, b, p(a-b), and a+b are pth powers modulo f.
Note that p is not precluded from being a pth power modulo f in Proposition (8). If [(ap+bp)/(a+b)] is a pth power and 2p does not divide a, b, a-b, or a+b, there is apparently nothing to prevent p from being a pth power modulo every prime factor of [(ap-bp)/(a-b)] not of the form p2k+1. By these propositions, if [(ap+bp)/(a+b)] is a pth power, 2p does not divide a, b, a-b, or a+b, and p is a pth power modulo every prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, then a, b, a-b, and [a+b] are not pth powers. (By Proposition (3), there would be at least one prime factor f of [(ap-bp)/(a-b)] of the form p2k+1 such that p was not a pth power modulo f. Then by Proposition (8), a, b, a-b, or [a+b] couldn't be a pth power.) This is the origin of the condition that a, b, a-b, or [a+b] be a pth power in the second conjecture; the desired condition where neither 2 or p is a pth power modulo a prime factor of [(ap-bp)/(a-b)] is then attained. More propositions are;
(9) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, and p is not a pth power modulo f, then (1) pa, p2b, p(a-b), and a+b are pth powers modulo f if 2 divides a, p does not divide a, and p/2 is a pth power modulo f, or (2) pa, b, a-b, and p2(a+b) are pth powers modulo f if 2 divides a, p does not divide a, and 2p is a pth power modulo f, or (3) p2a, pb, p(a-b), and a+b are pth powers modulo f if 2 divides b, p does not divide b, and p/2 is a pth power modulo f, or (4) a, pb, a-b, and p2(a+b) are pth powers modulo f if 2 divides b, p does not divide b, and 2p is a pth power modulo f.
(10) If [(ap+bp)/(a+b)] is a pth power, 2 divides a, p does not divide a, and f is a prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, then pa, 2pb, 22p2(a-b), and 22p(a+b) are pth powers modulo f. If [(ap+bp)/(a+b)] is a pth power, 2 divides b, p does not divide b, and f is a prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, then 2pa, pb, 22p2(a-b), and 22p(a+b) are pth powers modulo f.
Note that in Proposition (10), p is not precluded from being a pth power modulo f. Also, note that the power of p factors in Proposition (10) (where p is a factor of a, p is a factor of b, p2 is a factor of a-b, and p is a factor of a+b) are the same as those in Proposition (8) (where f is of the form p2k+1) when p divides a+b. Also, note that the power of p factors of a-b and a+b in Proposition (10) are different from those in Proposition (9) (where p or 1 is the factor of a-b, and 1 or p2 is the factor of a+b). This is not unexpected since f is not of the form p2k+1 in Proposition (9). Note that since Propositions (8), (9), and (10) are based solely on data collected for p=3, their form is sometimes ambiguous in that the p2 and 22 factors might be pp-1 and 2p-1 instead. If p(a+b)/2 is a pth power w.r.t. [(ap-bp)/(a-b)] , f is a prime factor of [(ap-bp)/(a-b)], and p/2 or 2p is a pth power modulo f, then a+b or p2(a+b) is a pth power modulo f, so the p2 factor of a+b in Proposition (9) is unambiguous. Propositions (9) and (10) are consistent when p is not a pth power modulo f only if 8 is a pth power modulo f, but 8 can be a pth power modulo f only if p=3 (2f-1≡1(mod f) and (23)(f-1)/p≡1(mod f), p≠3, implies 2(f-1)/p≡1(mod f) [since in this case, the greatest common divisor of f-1 and 3(f-1)/p is (f-1)/p], a contradiction). This follows from eliminating b, a-b, and a+b from the congruences if 2 divides a, or eliminating a, a-b, and a+b from the congruences if 2 divides b. Note that if a 2p-1p factor of the a+b term in Proposition (10) had been used, some inconsistency for p>3 could have been avoided. However, using a 2p-1pp-1, 22pp-1, 2p-1p2, or 22p2 factor of the a-b term in Proposition (10) implies that 8 is a pth power modulo f, a contradiction for p>3. Other propositions are;
(11) If [(ap+bp)/(a+b)] is a pth power, then a is a pth power w.r.t. (ap-bp)/(a-b) if 2p divides a, or b is a pth power w.r.t. (ap-bp)/(a-b) if 2p divides b, or p(a-b) and a+b are pth powers w.r.t. (ap-bp)/(a-b) if 2p divides a-b, or p2(a-b) and p(a+b) are pth powers w.r.t. (ap-bp)/(a-b) if 2p divides a+b.
(12) If [(ap+bp)/(a+b)] is a pth power, then pa is a pth power w.r.t. (ap-bp)/(a-b) if 2 divides a and p does not divide a, or pb is a pth power w.r.t. (ap-bp)/(a-b) if 2 divides b and p does not divide b.
(13) If [(ap+bp)/(a+b)] is a pth power and 2p divides a-b or a+b, then 2 is a pth power w.r.t. (ap-bp)/(a-b).
(14) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, and p is not a pth power modulo f, then (1) a, pb, a-b, and p2(a+b) are pth powers modulo f if 2p divides a and p/2 is a pth power modulo f, or (2) a, p2b, p2(a-b), and p(a+b) are pth powers modulo f if 2p divides a and 2p is a pth power modulo f, or (3) pa, b, a-b, and p2(a+b) are pth powers modulo f if 2p divides b and p/2 is a pth power modulo f, or (4) p2a, b, p2(a-b), and p(a+b) are pth powers modulo f if 2p divides b and 2p is a pth power modulo f, or (5) pa, p2b, p(a-b), and a+b, or p2a, pb, p(a-b), and a+b are pth powers modulo f if 2p divides a-b, or (6) a, p2b, p2(a-b), and p(a+b), or p2a, b, p2(a-b), and p(a+b) are pth powers modulo f if 2p divides a+b.
(15) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, and p is a pth power modulo f, then (1) a, 2b, 22(a-b), and 22(a+b) (and not 2) are pth powers modulo f is 2 divides a, or (2) 2a, b, 22(a-b), and 22(a+b) (and not 2) are pth powers modulo f if 2 divides b, or (3) a-b, a+b, and 2 (and not a or b) are pth powers modulo f if 2 divides a-b or a+b.
(16) If [(ap+bp)/(a+b)] is a pth power, 2p divides a or b, [(ap-bp)/(a-b)] has two distinct prime factors, and neither distinct prime factor is of the form p2k+1, then [(ap-bp)/(a-b)] is of the form p2k+1, and (1) 2p is a pth power modulo both distinct prime factors, or (2) p is a pth power modulo both distinct prime factors, or (3) p/2 is a pth power modulo both distinct prime factors.
In this section, it is assumed that [(ap-bp)/(a-b)] can't be a pth power when [(ap+bp)/(a+b)] is a pth power. (As will be shown, the congruence properties of the prime factors of [(ap+bp)/(a+b)] when [(ap+bp)/(a+b)] is a pth power are similar to the congruence properties of the prime factors of [(ap-bp)/(a-b)] when [(ap+bp)/(a+b)] is a pth power, but are not the same.)
Let T be a natural number. If p=3, every prime factor of T is of the form pk+1, and T has n such distinct prime factors, then Tp or pTp has exactly pn representations of the form (ap+bp)/(a+b). Proving the first conjecture when p divides a+b would entail proving that if one representation of pTp of the form (ap+bp)/(a+b) exists, then other representations exist and that 2 and p split for some of these representations. There is little evidence that there would exist different representations of pTp of the form (ap+bp)/(a+b) for p>3. Even if there were a representation ((a')p+(b')p)/(a'+b') with split 2 and p, how to deal with the case where p was a pth power modulo every prime factor of [((a')p-(b')p)/(a'-b')] not of the form p2k+1 is unknown.
(17) If [(ap-bp)/(a-b)] is prime, qp-1≡1(mod p2), and q divides a, b, a+b, or a-b, then q is a pth power modulo [(ap-bp)/(a-b)].
(18) If p>3, [(ap-bp)/(a-b)] is prime, and p2 divides a, b, a+b, or a-b, then p is a pth power modulo [(ap-bp)/(a-b)]. If p=3, [(ap-bp)/(a-b)] is prime, and p2 divides a, b, or a+b or p3 divides a-b, then p is a pth power modulo [(ap-bp)/(a-b)].
(19) If [(ap-bp)/(a-b)] is prime, p does not divide q, p divides a and q divides a, or p divides b and q divides b, or p divides a+b and q divides a+b or a-b, then q is a pth power modulo [(ap-bp)/(a-b)]. If p>3, [(ap-bp)/(a-b)] is prime, p does not divide q, p divides a-b and q divides a+b or a-b, then q is a pth power modulo [(ap-bp)/(a-b)]. If p=3, [(ap-bp)/(a-b)] is prime, p does not divide q, p2 divides a-b and q divides a+b or a-b, then q is a pth power modulo [(ap-bp)/(a-b)].
(20) If p>3, [(ap-bp)/(a-b)] is a prime of the form p2k+1, and p2 divides a, b, a+b, or a-b, then a, b, a+b, a-b, and p are pth powers modulo [(ap-bp)/(a-b)]. If p=3, [(ap-bp)/(a-b)] is a prime of the form p2k+1, and p2 divides a, b, or a+b or p3 divides a-b, then a, b, a+b, a-b, and p are pth powers modulo [(ap-bp)/(a-b)].
Propositions (17), (18), and (19) and Propositions (5) and (6) lead to the following proposition;
(21) If [(ap+bp)/(a+b)] is a pth power, 2p divides a, b, a-b, or a+b, and [(ap-bp)/(a-b)] is prime, then every factor of a, b, a-b, and a+b is a pth power modulo [(ap-bp)/(a-b)] (note that this implies [(ap-bp)/(a-b)] is a prime of the form p2k+1).
Based on data collected for p=3, generalized versions of Propositions (17), (18), and (19) are true when [(ap-bp)/(a-b)]=Uk where U is a prime and p does not divide k. (The propositions would be modified so that the modulus would be U instead of [(ap-bp)/(a-b)].) This gives the following proposition;
(22) If [(ap+bp)/(a+b)] is a pth power, 2p divides a, b, a-b, or a+b, and [(ap-bp)/(a-b)]=Uk where U is a prime and p does not divide k, then every factor of a, b, a-b, and a+b is a pth power modulo U.
(23) If p>3, qp-1≠1(mod p2), and q is a pth power w.r.t. (ap+bp)/(a+b), then p divides a if q divides a, p divides b if q divides b, or p divides a-b or a+b if q divides a-b or a+b. If p=3, qp-1≠1(mod p2), and q is a pth power w.r.t. (ap+bp)/(a+b), then p divides a-b or a+b if q divides a-b or a+b.
This proposition precludes first-case solutions of Fermat's equation except when 2p-1≡1(mod p2) or p=3 since ap+bp=cp implies cp+bp divides ap+2bp, cp+ap divides 2ap+bp, and ap-bp divides cp-2ap and hence that 2 is a pth power w.r.t. (cp+bp)/(c+b), (cp+ap)/(c+a), and (ap-bp)/(a-b). (If ap+bp=cp, one of a, b, and c must be even.) Mirimanoff proved that a first-case solution of Fermat's equation implies that 3p-1≡1(mod p2). The probability that a root of the congruence xp-1≡1(mod p2), 0<x<p2, is one larger than another root is 1/p (since there are p-1 roots having p(p-1) possible values). There then shouldn't be any p such that 3p-1≡2p-1≡1(mod p2) since the sum of (1/p)(1/p) over all p converges and the only p less than 3x109 such that 2p-1≡1(mod p2) are 1093 and 3511, and 3p-1≠1(mod p2) for either of these p.
Let ζ be a primitive pth root of unity and K=Q(ζ), a cyclotomic field of degree p-1 over Q. Let λ denote 1-ζ. The following proposition is based on data collected for p=3;
(24) q is a pth power w.r.t. (ap+bp)/(a+b) if and only if q is congruent to the pth power of an integer (in K) modulo a+ζb, a+ζ2b, a+ζ3b, ..., and a+ζp-1b.
Barlow's formulas implied by a second-case solution of Fermat's equation where p divides c are (cp-bp)/(c-b)=Rp, (cp-ap)/(c-a)=Sp, (ap+bp)/(a+b)=pTp, c-b=rp, c-a=sp, and a+b=(pkt)p/p where rR=a, sS=b, pktT=c, and g.c.d.(r, R)=g.c.d.(s, S)=g.c.d.(pkt, T)=1. Then c divides Rp-bp-1 and Sp-ap-1 and hence b(f-1)/p≡1(mod f) and a(f-1)/p≡1(mod f) for every prime factor f of T. Also, b divides pTp-ap-1 and Rp-cp-1 and hence (pa)(f-1)/p≡1(mod f) and c(f-1)/p≡1(mod f) for every prime factor f of S. (Analogous results hold for a.) There can be second-case solutions of Fermat's equation where p divides c only if c, pb, and c-b are pth powers w.r.t. (cp-bp)/(c-b), c, pa, and c-a are pth powers w.r.t. (cp-ap)/(c-a), a, b, and p(a+b) are pth powers w.r.t. (ap+bp)/(a+b), and every prime factor of (ap+bp)/(a+b)/p is of the form p2k+1 (based on these formulas, the prime factors of (cp-bp)/(c-b) and (cp-ap)/(c-a) are not necessarily of the form p2k+1). (The requirement that every prime factor of (ap+bp)/(a+b)/p be of the form p2k+1 could be said to be another characteristic property of the equation ap+bp=cp, p divides c.) Also, p divides ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2 so that bp-1≡Tp(mod a+b) (this is relevant to fractional ideals to be discussed in the next section). The following proposition is based on data collected for p=3, 5, 7, and 11;
(25) If p>3 and a is a pth power w.r.t. (ap+bp)/(a+b), then ap-1≡1(mod p2) if p does not divide a. If p=3, a is a pth power w.r.t. (ap+bp)/(a+b), and p divides b or a-b or p2 divides a+b, then ap-1≡1(mod p2). (Analogous results hold for b.) If a-b is a pth power w.r.t. (ap+bp)/(a+b), then (a-b)p-1≡1(mod p2) if p does not divide a-b or a+b. (Analogous results hold for a+b.)
Proposition (25) shows that a second-case solution of Fermat's equation where p divides c implies ap-1≡1(mod p2) and bp-1≡1(mod p2) (avoiding the constraint that p not divide c in Furtwängler's first theorem).
The question of whether a larger power of p divides (qp-1-1)/p than a+b can be avoided by considering the reciprocal of α. ((a+ζ2b)/(a+ζb))((a+ζ3b)/(a+ζ2b))···((a+ζpb)/(a+ζp-1b))=(a+b)/(a+ζb). Let d(m,n) denote m "choose" n (a binomial coefficient). Let f1 denote d(1,0)ap-2-d(2,1)ap-3b+d(3,2)ap-4b2-...-d(p-1,p-2)bp-2, f2 denote d(2,0)ap-3-d(3,1)ap-4b+d(4,2)ap-5b2-...+d(p-1,p-3)bp-3, f3 denote d(3,0)ap-4-d(4,1)ap-5b+d(5,2)ap-6b2-...-d(p-1,p-4)bp-4, ..., and fp-1 denote d(p-1,0). Collecting terms in the product (1-bλζ/(a+ζb))(1-bλζ2/(a+ζ2b))···(1-bλζp-1/(a+ζp-1b)) gives (a+b)/(a+ζb)=1+(bλf1+b2λ2f2+...+bp-1λp-1ff-1)/((ap+bp)/(a+b)). Tr(1)=p-1 and Tr(λk)=p. Set α to (a+b)/(a+ζb). Substituting for Tr(1), Tr(λ), Tr(λ2), ..., Tr(λp-2) and collecting terms gives Tr((α-1)/λ)=b((p-1)ap-2-(p-2)ap-3b+(p-3)ap-4b2-...-bp-2)/((ap+bp)/(a+b)). Setting β to qp-1 where q divides b gives {qp-1/[(a+b)/(a+ζb)]}=ζuv where v=b((p-1)ap-2-(p-2)ap-3b+(p-3)ap-4b2-...-bp-2)/((ap+bp)/(a+b)). If p divides a+b, then p also divides (ap+bp)/(a+b). (p-1)ap-2-(p-2)ap-3b+(p-3)ap-4b2-...-bp-2 is congruent to -(a+b)p-2 modulo p, therefore if p divides a+b, then 1/p does not divide v. By Proposition (25), if ap+bp=cp, p divides c, then bp-1≡1(mod p2) and hence {bp-1/[(a+b)/(a+ζb)]}=1.
p=(1-ζ)(1-ζ2)(1-ζ3)···(1-ζp-1) and the ideals [1-ζ], [1-ζ2], [1-ζ3], ..., [1-ζp-1] are equal. If p divides a+b and (ap+bp)/(a+b)/p is a pth power, then ((a+ζ2b)/(1-ζ2))/((a+ζb)/(1-ζ)), ((a+ζ3b)/(1-ζ3))/((a+ζ2b)/(1-ζ2)), ((a+ζ4b)/(1-ζ4)/((a+ζ3b)/(1-ζ3)), ..., ((a+ζp-1b)/(1-ζp-1))/((a+ζp-2b)/(1-ζp-2)) are pth power of fractional ideals and hence (a+ζ2b)/(a+ζb), (a+ζ3b)/(a+ζ2b), (a+ζ4b)/(a+ζ3b), ..., (a+ζp-1b)/(a+ζp-2b) are pth powers of fractional ideals. Furthermore, (a+ζi+1b)/(a+ζib)=1-bλζi/(a+ζib)=αi≡1(mod λ), i=1, 2, 3, ..., p-2, and hence {β/αi}=1 for all β in Q(ζ) that are prime to αi. Tr(-bζ/(a+ζb))+Tr(-bζ2/(a+ζ2b))+Tr(-bζ3/(a+ζ3b))+...+Tr(-bζp-1/(a+ζp-1b))=b(ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2)/((ap+bp)/(a+b)), therefore if p divides a+b and (ap+bp)/(a+b)/p is a pth power, then {qp-1/[(a+b)/(a+ζp-1b)]}=ζuw where w=b(ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2)/((ap+bp)/(a+b)). ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2 is congruent to (a+b)p-2 modulo p, therefore 1/p does not divide w. By Proposition (25), if ap+bp=cp, p divides c, then bp-1≡1(mod p2) and hence {bp-1/[(a+b)/(a+ζp-1b)]}=1. The following proposition is based on data collected for p=3;
(26) If 2 divides b, p divides a+b, and (ap+bp)/(a+b)/p is a pth power, then p3 divides (p-1)ap-2-(p-2)ap-3b+(p-3)ap-4b2-...-bp-2 and p2 does not divide ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2. If 2p divides a+b and (ap+bp)/(a+b)/p is a pth power, then p2 does not divide (p-1)ap-2-(p-2)ap-3b+(p-3)ap-4b2-...-bp-2 or ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2.
By Proposition (5), if 2p-1≠1(mod p2), 2 divides b, p divides a+b, and (ap+bp)/(a+b)/p is a pth power, then bp-1≠1(mod p2). Proposition (26) then implies that if 2p-1≠1(mod p2), 2 divides b, p divides a+b, and (ap+bp)/(a+b)/p is a pth power, then {bp-1/[(a+ζb)/(a+b)]}≠1, {bp-1/[(a+b)/(a+ζb)]}=1, and {bp-1/[(a+b)/(a+ζp-1b)]}≠1.
(27) If [(ap+bp)/(a+b)]=Tp and T=Uk where U is a prime and p does not divide k, 2p divides a, b, a-b, or a+b, 2 does not divide a, q divides a, and qp-1≡1(mod p3), then q is a pth power w.r.t. (ap+bp)/(a+b). If [(ap+bp)/(a+b)] is a pth power, 2p divides a, b, a-b, or a+b, 2 does not divide a, and q divides a, then q is a pth power w.r.t. (ap+bp)/(a+b) only if qp-1≡1(mod p3). (Analogous results hold for b, a-b, and a+b.) If [(ap+bp)/(a+b)]=Tp and T=Uk where U is a prime and p does not divide k, 2 divides a, p does not divide a, q divides a, and every prime factor of q is a pth power modulo p2, then q is a pth power w.r.t. (ap+bp)/(a+b). (Analogous results hold for b.) If [(ap+bp)/(a+b)]=Tp and T=Uk where U is a prime and p does not divide k, p divides a+b, 2 does not divide a+b, q divides a+b, and qp-1≡1(mod p3), then q is a pth power w.r.t. (ap+bp)/(a+b). If [(ap+bp)/(a+b)] is a pth power, p divides a+b, 2 does not divide a+b, q divides a+b, and p does not divide q, then q is a pth power w.r.t. (ap+bp)/(a+b) only if qp-1≡1(mod p3).
(28) If [(ap+bp)/(a+b)]=Tp where every prime factor of T is of the form p2k+1, p3 divides a, b, or a-b or p4 divides a+b, and 2 does not divide a, then ap-1≡1(mod p3). If [(ap+bp)/(a+b)]=Tp where T has only one distinct prime factor, this prime factor is of the form p2k+1, p3 divides a', b', or a'-b' or p4 divides a'+b' for some representation [((a')p+(b')p)/(a'+b')] of Tp, and 2 does not divide a, then ap-1≡1(mod p3). (Analogous results hold for b and a-b.) If [(ap+bp)/(a+b)]=Tp where every prime factor of T is of the form p2k+1, p3 divides a, b, or a-b or p4 divides a+b, and 2 does not divide a+b, then [(a+b)/p]p-1≡1(mod p3) if p divides a+b, or (a+b)p-1≡1(mod p3) if p does not divide a+b. If [(ap+bp)/(a+b)]=Tp where T has only one distinct prime factor, this prime factor is of the form p2k+1, p3 divides a', b', or a'-b' or p4 divides a'+b' for some representation [((a')p+(b')p)/(a'+b')] of Tp, and 2 does not divide a+b, then [(a+b)/p]p-1≡1(mod p3) if p divides a+b, or (a+b)p-1≡1(mod p3) if p does not divide a+b. If [(ap+bp)/(a+b)]=Tp where every prime factor of T is of the form p2k+1, 2 divides a, and p does not divide a, then (a/2)p-1≡1(mod p3). (Analogous results hold for b.) If [(ap+bp)/(a+b)]=Tp where every prime factor of T is of the form p2k+1, 2 does not divide a, and p divides a+b, then ap-1≡1(mod p3). (Analogous results hold for b and a-b.)
(29) If [(ap+bp)/(a+b)]=Tp and T=Uk where U is a prime and p does not divide k, then p is a pth power w.r.t. (ap+bp)/(a+b) if and only if p3 divides a', b', or a'-b' or p4 divides a'+b' for some representation [((a')p+(b')p)/(a'+b')] of Tp. If [(ap+bp)/(a+b)]=Tp and T=Uk where U is a prime and p divides k, then p is not a pth power w.r.t. (ap+bp)/(a+b).
(30) If [(ap+bp)/(a+b)] is a pth power, then pp-1a is a pth power w.r.t. (ap+bp)/(a+b) if 2p divides a, pp-1b is a pth power w.r.t. (ap+bp)/(a+b) if 2p divides b, pp-1(a-b) and a+b are pth powers w.r.t. (ap+bp)/(a+b) if 2p divides a-b, or a-b and p(a+b) are pth powers w.r.t. (ap+bp)/(a+b) if 2p divides a+b.
Note that ap+bp=cp, p divides c, implies that p(a+b) is a pth power w.r.t. (ap+bp)/(a+b) and also that 2p divides a+b (if Proposition (5) is accepted). There is no apparent reason (other than the "difficult proof" hypothesis) why p(a+b) should be a pth power w.r.t. (ap+bp)/(a+b) when [(ap+bp)/(a+b)] is a pth power and 2p divides a+b.
(31) If [(ap+bp)/(a+b)] is a pth power, then a is a pth power w.r.t. (ap+bp)/(a+b) if 2 divides a and p does not divide a, or b is a pth power w.r.t. (ap+bp)/(a+b) if 2 divides b and p does not divide b.
(32) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap+bp)/(a+b)] of the form p2k+1, and p is not a pth power modulo f, then (1) pp-1a, pp-1b, pp-1(a-b), and a+b are pth powers modulo f if p divides a, b, or a-b, or (2) a, b, a-b, and p(a+b) are pth powers modulo f if p divides a+b. If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap+bp)/(a+b)] of the form p2k+1, and p is a pth power modulo f, then a, b, a-b, and a+b are pth powers modulo f.
Note that ap+bp=cp, p divides c, implies every prime factor of [(ap+bp)/(a+b)] is of the form p2k+1 and also that a and b are pth powers w.r.t. (ap+bp)/(a+b). There is no apparent reason (other than the "difficult proof" hypothesis) why a and b should be pth powers modulo f when [(ap+bp)/(a+b)] is a pth power, p divides a+b, and f is a prime factor of [(ap+bp)/(a+b)] of the form p2k+1.
(33) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap+bp)/(a+b)] not of the form p2k+1, and p is not a pth power modulo f, then (1) pp-1a, b, p(a-b), and pp-1(a+b), or pp-1a, pb, a-b, and p(a+b) are pth powers modulo f if 2p divides a, or (2) a, pp-1b, p(a-b), and pp-1(a+b), or pa, pp-1b, a-b, and p(a+b) are pth powers modulo f if 2p divides b, or (3) a, pb, pp-1(a-b), and a+b, or pa, b, pp-1(a-b), and a+b are pth powers modulo f if 2p divides a-b, or (4) pa, pp-1b, a-b, and p(a+b), or pp-1a, pb, a-b, and p(a+b) are pth powers modulo f if 2p divides a+b.
(34) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap+bp)/(a+b)] not of the form p2k+1, and p is not a pth power modulo f, then (1) a, pb, pp-1(a-b), and a+b, or a, pp-1b, p(a-b), and pp-1(a+b) are pth powers modulo f if 2 divides a and p does not divide a, or (2) pa, b, pp-1(a-b), and a+b, or pp-1a, b, p(a-b), and pp-1(a+b) are pth powers modulo f if 2 divides b and p does not divide b.
Note that since Propositions (30), (32), (33), and (34) are based solely on data collected for p=3, their form is ambiguous in that the p exponents might be 2 instead of p-1. Congruence properties of the prime factors of [(ap-bp)/(a-b)] when [(ap+bp)/(a+b)] is a pth power appear to determine the form of Propositions (30) through (34). (Propositions (30), (31), (32), (33), and (34) can be transformed into Propositions (11), (12), (8), (14), and (9) respectively by multiplying the a, b, a-b, and a+b terms by p and switching the a+b and a-b terms [and of course the modulo bases]. This is just an attempt to find a simple relationship between the congruence properties of the prime factors of [(ap+bp)/(a+b)] and [(ap-bp)/(a-b)] when [(ap+bp)/(a+b)] is a pth power and has no apparent logical basis.)
(35) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap+bp)/(a+b)] not of the form p2k+1, and p is a pth power modulo f, then (1) a (and not b, a-b, or a+b) is a pth power modulo f if 2 divides a, or (2) b (and not a, a-b, or a+b) is a pth power modulo f if 2 divides b, or (3) a-b and a+b (and not a or b) are pth powers modulo f if 2 divides a-b or a+b.
As shown previously, ap+bp=cp, p divides c, implies c, pb, and c-b are pth powers w.r.t. (cp-bp)/(c-b), c, pa, and c-a are pth powers w.r.t. (cp-ap)/(c-a), and 2 and p are common factors of c. Then by Propositions (33) and (35), every prime factor of (cp-bp)/(c-b) and (cp-ap)/(c-a) must be of the form p2k+1. (Substituting c for a and -b for b in Proposition (33) gives pp-1c, -b, p(c+b), and pp-1(c-b), or pp-1c, -pb, c+b, and p(c-b) are pth powers modulo f [a prime factor of (cp-bp)/(c-b)] if 2p divides c and p is not a pth power modulo f [a contradiction]. Substituting c for a and -b for b in Proposition (35) gives c [and not c-b] is a pth power modulo f if 2p divides c and p is a pth power modulo f [a contradiction]. Similar results follow by substituting c for a and -a for b in Propositions (33) and (35).) Furthermore, by Proposition (32), p must be a pth power w.r.t. (cp-bp)/(c-b) and (cp-ap)/(c-a). As shown previously, ap+bp=cp, p divides c, implies a, b, and p(a+b) are pth powers w.r.t. (ap+bp)/(a+b), and every prime factor of [(ap+bp)/(a+b)] is of the form p2k+1. This gives the following proposition;
(36) If ap+bp=cp where p divides c, then every prime factor of (cp-bp)/(c-b) is of the form p2k+1 and c, b, c-b, c+b, and p are pth powers w.r.t. (cp-bp)/(c-b). (Analogous results hold for (cp-ap)/(c-a).) If ap+bp=cp where p divides c, then every prime factor of [(ap+bp)/(a+b)] is of the form p2k+1 and a, b, a-b, and p(a+b) are pth powers w.r.t. (ap+bp)/(a+b).
Further evidence for the above proposition is given by the following proposition (based on data collected for p=3, 5, 7, and 11);
(37) If every prime factor of [(ap+bp)/(a+b)] is of the form p2k+1, and p2 divides a, b, a-b, or a+b, then ap-1≡1(mod p2) if p does not divide a, bp-1≡1(mod p2) if p does not divide b, and (a-b)p-1≡1(mod p2) and (a+b)p-1≡1(mod p2) if p does not divide a-b or a+b.
Substituting c for a and -b for b in Proposition (14) gives p2(c-b) is a pth power modulo f (a prime factor of (cp+bp)/(c+b) not of the form p2k+1) if p and 2p are not pth powers modulo f, or p(c-b) is a pth power modulo f if p is not a pth power modulo f and 2p is a pth power modulo f. Then if c-b is a pth power, p must be a pth power modulo f and hence by Proposition (7), 2 cannot be a pth power modulo f (otherwise, 2p would be a pth power modulo f). As shown previously, ap+bp=cp implies 2 is a pth power w.r.t. (cp+bp)/(c+b). This gives the following proposition;
(38) If ap+bp=cp where p divides c, then every prime factor of (cp+bp)/(c+b) is of the form p2k+1 and c, b, p(c+b) and c-b are pth powers w.r.t. (cp+bp)/(c+b). (Analogous results follow for (cp+ap)/(c+a).)
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Bericht über neuere Undersuchungen und Probleme aus der Theorie der algebraischen Zahlkörper. Teil I: Klassenkörperietheorie. Teil Ia: Beweise zu Tiel I; Teil II: Reziprozitätsgesetz. 3. Aufl.. Wurzburg Wien: Physika-Verlag. 204 pp. Jbuch 52-150; 53, 143; 46-165
Variants of Fermat's Last Theorem
Let a, b, and c be natural numbers relatively prime in pairs and let p be an odd prime. Every prime factor of (ap+bp)/(a+b) other than p is of the form pk+1. p (and no higher power of p) divides (ap+bp)/(a+b) if and only if p divides a+b. Let q be a natural number. q will be said to be a pth power with respect to (ap+bp)/(a+b) if q(f-1)/p≡1(mod f) for every prime factor f, f≠p, of (ap+bp)/(a+b). Let [(ap+bp)/(a+b)] denote (ap+bp)/(a+b)/p if p divides a+b, or (ap+bp)/(a+b) otherwise. Similarly, let [a+b] denote (a+b)/p if p divides a+b, or a+b otherwise. The following two conjectures are the main topic of this article;(1) If p>3, there do not exist a and b such that [(ap+bp)/(a+b)] is a pth power.
(2) If p>3, there do not exist a and b such that [(ap+bp)/(a+b)] is a pth power, 2p does not divide a, b, a-b, or a+b, and a, b, a-b, or [a+b] is a pth power.
Note that if the first conjecture is true, there are no solutions of Fermat's equation ap+bp=cp (which, of course, is already known), and that the second conjecture encompasses the first case of Fermat's Last Theorem (where p does not divide abc). (In 1810, Barlow proved that ap+bp=cp only if [(ap+bp)/(a+b)] is a pth power.) Let T be a natural number. Since a, b, and T are not symmetrical in the equation [(ap+bp)/(a+b)]=Tp, it is not obvious how to apply the theory of elliptic curves to these problems. The "pth power w.r.t." concept and the identity ap+bp=(ap-bp)+2bp play a central role in proving these conjectures. For example, if p divides a+b and (ap+bp)/(a+b)/p is a pth power, then p(a+b)/2 is a pth power w.r.t. (ap-bp)/(a-b) (this is a succinct way of saying that (p(a+b))(f-1)/p≡2(f-1)/p(mod f) for every prime factor f, f≠p, of (ap-bp)/(a-b)). Note that if p is not a pth power modulo a prime f of the form pk+1, then, for example, if f does not divide a+b, (px(a+b))(f-1)/p≡1(mod f), 0≤x<p, has a solution (x defines a congruence class). Furthermore, if 2 is not a pth power modulo f, then, for example, (2yp2(a+b))(f-1)/p≡1(mod f), 0≤y<p, has a solution and a+b can be eliminated from the congruences. The objective in the following is to eliminate a, b, a-b, and a+b from certain congruences so that congruence relationships involving only 2 and p are obtained. When p=3, there do exist a and b such that [(ap+bp)/(a+b)] is a pth power and many properties of such a and b, among them reformulated versions of the classical Furtwängler and Vandiver theorems for Fermat's equation, can be empirically derived. In the following, these "propositions" are stated as if they were true for all p. One justification for doing this is the first conjecture above. Another justification for doing this for some of the propositions is the "difficult proof" hypothesis (not rigorous, of course). This hypothesis is; a, b, and T satisfying the equation [(ap+bp)/(a+b)]=Tp and having characteristic properties of the equation ap+bp=cp will have other properties in common with a, b, and c satisfying the equation ap+bp=cp (otherwise, there would be a contradiction and there would be a "simple" proof of Fermat's Last Theorem). For example, it can be easily proved that ap+bp=cp implies 2 is a pth power w.r.t. (ap-bp)/(a-b). Based on empirical evidence collected for p=3, if [(ap+bp)/(a+b)] is a pth power and 2p divides a+b, then 2 is a pth power w.r.t. (ap-bp)/(a-b) (although there is no apparent reason why this should be true [other than the "difficult proof" hypothesis]). More will be said on what these characteristic properties are later.
Congruence Properties of Prime Factors of [(ap-bp)/(a-b)] when [(ap+bp)/(a+b)] is a Pth Power
The following propositions are based on data collected for p=3;(3) If [(ap+bp)/(a+b)] is a pth power and 2p does not divide a, b, a-b, or a+b, then 2 and p are not pth powers w.r.t. (ap-bp)/(a-b).
(4) If [(ap+bp)/(a+b)] is a pth power, 2p does not divide a, b, a-b, or a+b, and f is a prime factor of [(ap-bp)/(a-b)], then 2 is a pth power modulo f if and only if f is of the form p2k+1.
By these two propositions, if [(ap+bp)/(a+b)] is a pth power and 2p does not divide a, b, a-b, or a+b, then there is at least one prime factor of [(ap-bp)/(a-b)] not of the form p2k+1. In 1912, Furtwängler proved that if ap+bp=cp, q divides a and p does not divide ac, or q divides b and p does not divide bc, then qp-1≡1(mod p2). (Proofs of this theorem use the condition that a+b must be a pth power and it is not obvious how to prove a reformulated version of the theorem without using this condition.) Furtwängler also proved that if ap+bp=cp, q divides a-b or a+b, and p does not divide a-b or a+b, then qp-1≡1(mod p2). Note that if p does not divide a natural number d, then dp(p-1)≡1(mod p2) by Euler's theorem. Then if qp-1≡1(mod p2), q is a pth power modulo p2. The reformulated version of Furtwängler's theorems is;
(5) If [(ap+bp)/(a+b)] is a pth power and 2 does not divide a, then p does not divide a and every prime factor of a is a pth power modulo p2. If [(ap+bp)/(a+b)] is a pth power, 2 divides a, and p does not divide a, then a/2 is a pth power modulo p2. (Analogous results hold for b.) If [(ap+bp)/(a+b)] is a pth power and 2 does not divide a-b, then p does not divide a-b and every prime factor of a-b is a pth power modulo p2. If [(ap+bp)/(a+b)] is a pth power and 2 does not divide a+b, then p2 does not divide a+b and every prime factor of a+b other than p (if p divides a+b) is a pth power modulo p2.
The peculiar form of Furtwängler's second theorem, that is, the condition that p not divide a-b or a+b, makes sense when viewed from this perspective; 2 divides a-b if and only if 2 divides a+b. This proposition implies that if [(ap+bp)/(a+b)] is a pth power, p divides a, b, a-b, or a+b, and 2p does not divide a, b, a-b, or a+b, then 2 divides a or b, and p divides a+b. Note that this proposition implies "split" 2 and p are not possible when ap+bp=cp, p divides c, since p2 cannot divide a+b in this case. (By Barlow's formulas, p(a+b) must be a pth power when ap+bp=cp, p divides c.) The requirement that 2p divide a+b could be said to be a characteristic property of the equation ap+bp=cp, p divides c. A related proposition is;
(6) If [(ap+bp)/(a+b)] is a pth power, then p2 divides a if 2p divides a, p2 divides b if 2p divides b, p2 divides a-b if 2p divides a-b, or p3 divides a+b if 2p divides a+b.
In the following, p/2 will be said to be a pth power modulo a prime f if p(f-1)/p≡2(f-1)/p(mod f). More propositions are;
(7) If [(ap+bp)/(a+b)] is a pth power, 2 divides a or b, and f is a prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, then exactly one of 2p, p, or p/2 is a pth power modulo f.
(8) If [(ap+bp)/(a+b)] is a pth power, p divides a+b, and f is a prime factor of [(ap-bp)/(a-b)] of the form p2k+1, then pa, pb, p2(a-b), and p(a+b) are pth powers modulo f. If [(ap+bp)/(a+b)] is a pth power, p divides a, b, or a-b, and f is a prime factor of [(ap-bp)/(a-b)] of the form p2k+1, then a, b, p(a-b), and a+b are pth powers modulo f.
Note that p is not precluded from being a pth power modulo f in Proposition (8). If [(ap+bp)/(a+b)] is a pth power and 2p does not divide a, b, a-b, or a+b, there is apparently nothing to prevent p from being a pth power modulo every prime factor of [(ap-bp)/(a-b)] not of the form p2k+1. By these propositions, if [(ap+bp)/(a+b)] is a pth power, 2p does not divide a, b, a-b, or a+b, and p is a pth power modulo every prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, then a, b, a-b, and [a+b] are not pth powers. (By Proposition (3), there would be at least one prime factor f of [(ap-bp)/(a-b)] of the form p2k+1 such that p was not a pth power modulo f. Then by Proposition (8), a, b, a-b, or [a+b] couldn't be a pth power.) This is the origin of the condition that a, b, a-b, or [a+b] be a pth power in the second conjecture; the desired condition where neither 2 or p is a pth power modulo a prime factor of [(ap-bp)/(a-b)] is then attained. More propositions are;
(9) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, and p is not a pth power modulo f, then (1) pa, p2b, p(a-b), and a+b are pth powers modulo f if 2 divides a, p does not divide a, and p/2 is a pth power modulo f, or (2) pa, b, a-b, and p2(a+b) are pth powers modulo f if 2 divides a, p does not divide a, and 2p is a pth power modulo f, or (3) p2a, pb, p(a-b), and a+b are pth powers modulo f if 2 divides b, p does not divide b, and p/2 is a pth power modulo f, or (4) a, pb, a-b, and p2(a+b) are pth powers modulo f if 2 divides b, p does not divide b, and 2p is a pth power modulo f.
(10) If [(ap+bp)/(a+b)] is a pth power, 2 divides a, p does not divide a, and f is a prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, then pa, 2pb, 22p2(a-b), and 22p(a+b) are pth powers modulo f. If [(ap+bp)/(a+b)] is a pth power, 2 divides b, p does not divide b, and f is a prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, then 2pa, pb, 22p2(a-b), and 22p(a+b) are pth powers modulo f.
Note that in Proposition (10), p is not precluded from being a pth power modulo f. Also, note that the power of p factors in Proposition (10) (where p is a factor of a, p is a factor of b, p2 is a factor of a-b, and p is a factor of a+b) are the same as those in Proposition (8) (where f is of the form p2k+1) when p divides a+b. Also, note that the power of p factors of a-b and a+b in Proposition (10) are different from those in Proposition (9) (where p or 1 is the factor of a-b, and 1 or p2 is the factor of a+b). This is not unexpected since f is not of the form p2k+1 in Proposition (9). Note that since Propositions (8), (9), and (10) are based solely on data collected for p=3, their form is sometimes ambiguous in that the p2 and 22 factors might be pp-1 and 2p-1 instead. If p(a+b)/2 is a pth power w.r.t. [(ap-bp)/(a-b)] , f is a prime factor of [(ap-bp)/(a-b)], and p/2 or 2p is a pth power modulo f, then a+b or p2(a+b) is a pth power modulo f, so the p2 factor of a+b in Proposition (9) is unambiguous. Propositions (9) and (10) are consistent when p is not a pth power modulo f only if 8 is a pth power modulo f, but 8 can be a pth power modulo f only if p=3 (2f-1≡1(mod f) and (23)(f-1)/p≡1(mod f), p≠3, implies 2(f-1)/p≡1(mod f) [since in this case, the greatest common divisor of f-1 and 3(f-1)/p is (f-1)/p], a contradiction). This follows from eliminating b, a-b, and a+b from the congruences if 2 divides a, or eliminating a, a-b, and a+b from the congruences if 2 divides b. Note that if a 2p-1p factor of the a+b term in Proposition (10) had been used, some inconsistency for p>3 could have been avoided. However, using a 2p-1pp-1, 22pp-1, 2p-1p2, or 22p2 factor of the a-b term in Proposition (10) implies that 8 is a pth power modulo f, a contradiction for p>3. Other propositions are;
(11) If [(ap+bp)/(a+b)] is a pth power, then a is a pth power w.r.t. (ap-bp)/(a-b) if 2p divides a, or b is a pth power w.r.t. (ap-bp)/(a-b) if 2p divides b, or p(a-b) and a+b are pth powers w.r.t. (ap-bp)/(a-b) if 2p divides a-b, or p2(a-b) and p(a+b) are pth powers w.r.t. (ap-bp)/(a-b) if 2p divides a+b.
(12) If [(ap+bp)/(a+b)] is a pth power, then pa is a pth power w.r.t. (ap-bp)/(a-b) if 2 divides a and p does not divide a, or pb is a pth power w.r.t. (ap-bp)/(a-b) if 2 divides b and p does not divide b.
(13) If [(ap+bp)/(a+b)] is a pth power and 2p divides a-b or a+b, then 2 is a pth power w.r.t. (ap-bp)/(a-b).
(14) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, and p is not a pth power modulo f, then (1) a, pb, a-b, and p2(a+b) are pth powers modulo f if 2p divides a and p/2 is a pth power modulo f, or (2) a, p2b, p2(a-b), and p(a+b) are pth powers modulo f if 2p divides a and 2p is a pth power modulo f, or (3) pa, b, a-b, and p2(a+b) are pth powers modulo f if 2p divides b and p/2 is a pth power modulo f, or (4) p2a, b, p2(a-b), and p(a+b) are pth powers modulo f if 2p divides b and 2p is a pth power modulo f, or (5) pa, p2b, p(a-b), and a+b, or p2a, pb, p(a-b), and a+b are pth powers modulo f if 2p divides a-b, or (6) a, p2b, p2(a-b), and p(a+b), or p2a, b, p2(a-b), and p(a+b) are pth powers modulo f if 2p divides a+b.
(15) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap-bp)/(a-b)] not of the form p2k+1, and p is a pth power modulo f, then (1) a, 2b, 22(a-b), and 22(a+b) (and not 2) are pth powers modulo f is 2 divides a, or (2) 2a, b, 22(a-b), and 22(a+b) (and not 2) are pth powers modulo f if 2 divides b, or (3) a-b, a+b, and 2 (and not a or b) are pth powers modulo f if 2 divides a-b or a+b.
(16) If [(ap+bp)/(a+b)] is a pth power, 2p divides a or b, [(ap-bp)/(a-b)] has two distinct prime factors, and neither distinct prime factor is of the form p2k+1, then [(ap-bp)/(a-b)] is of the form p2k+1, and (1) 2p is a pth power modulo both distinct prime factors, or (2) p is a pth power modulo both distinct prime factors, or (3) p/2 is a pth power modulo both distinct prime factors.
In this section, it is assumed that [(ap-bp)/(a-b)] can't be a pth power when [(ap+bp)/(a+b)] is a pth power. (As will be shown, the congruence properties of the prime factors of [(ap+bp)/(a+b)] when [(ap+bp)/(a+b)] is a pth power are similar to the congruence properties of the prime factors of [(ap-bp)/(a-b)] when [(ap+bp)/(a+b)] is a pth power, but are not the same.)
Let T be a natural number. If p=3, every prime factor of T is of the form pk+1, and T has n such distinct prime factors, then Tp or pTp has exactly pn representations of the form (ap+bp)/(a+b). Proving the first conjecture when p divides a+b would entail proving that if one representation of pTp of the form (ap+bp)/(a+b) exists, then other representations exist and that 2 and p split for some of these representations. There is little evidence that there would exist different representations of pTp of the form (ap+bp)/(a+b) for p>3. Even if there were a representation ((a')p+(b')p)/(a'+b') with split 2 and p, how to deal with the case where p was a pth power modulo every prime factor of [((a')p-(b')p)/(a'-b')] not of the form p2k+1 is unknown.
Congruence Properties of Prime [(ap-bp)/(a-b)] when [(ap+bp)/(a+b)] is a Pth Power
In this section, more empirical evidence in support of Propositions (8), (11), and (13) is given. The following propositions are based on data collected for p=3, 5, 7, and 11;(17) If [(ap-bp)/(a-b)] is prime, qp-1≡1(mod p2), and q divides a, b, a+b, or a-b, then q is a pth power modulo [(ap-bp)/(a-b)].
(18) If p>3, [(ap-bp)/(a-b)] is prime, and p2 divides a, b, a+b, or a-b, then p is a pth power modulo [(ap-bp)/(a-b)]. If p=3, [(ap-bp)/(a-b)] is prime, and p2 divides a, b, or a+b or p3 divides a-b, then p is a pth power modulo [(ap-bp)/(a-b)].
(19) If [(ap-bp)/(a-b)] is prime, p does not divide q, p divides a and q divides a, or p divides b and q divides b, or p divides a+b and q divides a+b or a-b, then q is a pth power modulo [(ap-bp)/(a-b)]. If p>3, [(ap-bp)/(a-b)] is prime, p does not divide q, p divides a-b and q divides a+b or a-b, then q is a pth power modulo [(ap-bp)/(a-b)]. If p=3, [(ap-bp)/(a-b)] is prime, p does not divide q, p2 divides a-b and q divides a+b or a-b, then q is a pth power modulo [(ap-bp)/(a-b)].
(20) If p>3, [(ap-bp)/(a-b)] is a prime of the form p2k+1, and p2 divides a, b, a+b, or a-b, then a, b, a+b, a-b, and p are pth powers modulo [(ap-bp)/(a-b)]. If p=3, [(ap-bp)/(a-b)] is a prime of the form p2k+1, and p2 divides a, b, or a+b or p3 divides a-b, then a, b, a+b, a-b, and p are pth powers modulo [(ap-bp)/(a-b)].
Propositions (17), (18), and (19) and Propositions (5) and (6) lead to the following proposition;
(21) If [(ap+bp)/(a+b)] is a pth power, 2p divides a, b, a-b, or a+b, and [(ap-bp)/(a-b)] is prime, then every factor of a, b, a-b, and a+b is a pth power modulo [(ap-bp)/(a-b)] (note that this implies [(ap-bp)/(a-b)] is a prime of the form p2k+1).
Based on data collected for p=3, generalized versions of Propositions (17), (18), and (19) are true when [(ap-bp)/(a-b)]=Uk where U is a prime and p does not divide k. (The propositions would be modified so that the modulus would be U instead of [(ap-bp)/(a-b)].) This gives the following proposition;
(22) If [(ap+bp)/(a+b)] is a pth power, 2p divides a, b, a-b, or a+b, and [(ap-bp)/(a-b)]=Uk where U is a prime and p does not divide k, then every factor of a, b, a-b, and a+b is a pth power modulo U.
Wieferich's Criterion and the "Pth Power with Respect to" Concept
In 1909, Wieferich proved that if ap+bp=cp, p does not divide abc, then 2p-1≡1(mod p2). Wieferich derived this criterion from very complicated formulas; a simpler approach is to employ the "pth power w.r.t." concept. The following proposition is based on data collected for p=3, 5, 7, and 11;(23) If p>3, qp-1≠1(mod p2), and q is a pth power w.r.t. (ap+bp)/(a+b), then p divides a if q divides a, p divides b if q divides b, or p divides a-b or a+b if q divides a-b or a+b. If p=3, qp-1≠1(mod p2), and q is a pth power w.r.t. (ap+bp)/(a+b), then p divides a-b or a+b if q divides a-b or a+b.
This proposition precludes first-case solutions of Fermat's equation except when 2p-1≡1(mod p2) or p=3 since ap+bp=cp implies cp+bp divides ap+2bp, cp+ap divides 2ap+bp, and ap-bp divides cp-2ap and hence that 2 is a pth power w.r.t. (cp+bp)/(c+b), (cp+ap)/(c+a), and (ap-bp)/(a-b). (If ap+bp=cp, one of a, b, and c must be even.) Mirimanoff proved that a first-case solution of Fermat's equation implies that 3p-1≡1(mod p2). The probability that a root of the congruence xp-1≡1(mod p2), 0<x<p2, is one larger than another root is 1/p (since there are p-1 roots having p(p-1) possible values). There then shouldn't be any p such that 3p-1≡2p-1≡1(mod p2) since the sum of (1/p)(1/p) over all p converges and the only p less than 3x109 such that 2p-1≡1(mod p2) are 1093 and 3511, and 3p-1≠1(mod p2) for either of these p.
Let ζ be a primitive pth root of unity and K=Q(ζ), a cyclotomic field of degree p-1 over Q. Let λ denote 1-ζ. The following proposition is based on data collected for p=3;
(24) q is a pth power w.r.t. (ap+bp)/(a+b) if and only if q is congruent to the pth power of an integer (in K) modulo a+ζb, a+ζ2b, a+ζ3b, ..., and a+ζp-1b.
Barlow's Formulas and the "Pth Power with Respect to" Concept
Barlow's formulas implied by a first-case solution of Fermat's equation are (cp-bp)/(c-b)=Rp, (cp-ap)/(c-a)=Sp, (ap+bp)/(a+b)=Tp, c-b=rp, c-a=sp, and a+b=tp where rR=a, sS=b, tT=c, and g.c.d.(r, R)=g.c.d.(s, S)=g.c.d.(t, T)=1. Then a divides Sp-cp-1 and Tp-bp-1 and hence c(f-1)/p≡1(mod f) and b(f-1)/p≡1(mod f) for every prime factor f of R. (Note that c(f-1)/p≡b(f-1)/p(mod f) and cp≡bp(mod f) so that every prime factor of (cp-bp)/(c-b) must be of the form p2k+1 [first proved by Sophie Germain].) Analogous results hold for b and c. There can be first-case solutions of Fermat's equation only if c, b, and c-b are pth powers w.r.t. (cp-bp)/(c-b), c, a, and c-a are pth powers w.r.t. (cp-ap)/(c-a), a, b, and a+b are pth powers w.r.t. (ap+bp)/(a+b), and every prime factor of (cp-bp)/(c-b), (cp-ap)/(c-a), and (ap+bp)/(a+b) is of the form p2k+1. Also, r divides sp-tp, s divides rp-tp, and t divides rp+sp. If n>2, [(xn+yn)1/n-x]1/n+[(xn+yn)1/n-y]1/n>(x+y)1/n where x and y are positive real numbers, therefore r+s>t>r,s and hence each of r, s, and t has a prime factor of the form pk+1. Also, (ap+bp)/(a+b)=(a+b)[ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2]+pbp-1 so that pbp-1≡Tp(mod a+b).Barlow's formulas implied by a second-case solution of Fermat's equation where p divides c are (cp-bp)/(c-b)=Rp, (cp-ap)/(c-a)=Sp, (ap+bp)/(a+b)=pTp, c-b=rp, c-a=sp, and a+b=(pkt)p/p where rR=a, sS=b, pktT=c, and g.c.d.(r, R)=g.c.d.(s, S)=g.c.d.(pkt, T)=1. Then c divides Rp-bp-1 and Sp-ap-1 and hence b(f-1)/p≡1(mod f) and a(f-1)/p≡1(mod f) for every prime factor f of T. Also, b divides pTp-ap-1 and Rp-cp-1 and hence (pa)(f-1)/p≡1(mod f) and c(f-1)/p≡1(mod f) for every prime factor f of S. (Analogous results hold for a.) There can be second-case solutions of Fermat's equation where p divides c only if c, pb, and c-b are pth powers w.r.t. (cp-bp)/(c-b), c, pa, and c-a are pth powers w.r.t. (cp-ap)/(c-a), a, b, and p(a+b) are pth powers w.r.t. (ap+bp)/(a+b), and every prime factor of (ap+bp)/(a+b)/p is of the form p2k+1 (based on these formulas, the prime factors of (cp-bp)/(c-b) and (cp-ap)/(c-a) are not necessarily of the form p2k+1). (The requirement that every prime factor of (ap+bp)/(a+b)/p be of the form p2k+1 could be said to be another characteristic property of the equation ap+bp=cp, p divides c.) Also, p divides ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2 so that bp-1≡Tp(mod a+b) (this is relevant to fractional ideals to be discussed in the next section). The following proposition is based on data collected for p=3, 5, 7, and 11;
(25) If p>3 and a is a pth power w.r.t. (ap+bp)/(a+b), then ap-1≡1(mod p2) if p does not divide a. If p=3, a is a pth power w.r.t. (ap+bp)/(a+b), and p divides b or a-b or p2 divides a+b, then ap-1≡1(mod p2). (Analogous results hold for b.) If a-b is a pth power w.r.t. (ap+bp)/(a+b), then (a-b)p-1≡1(mod p2) if p does not divide a-b or a+b. (Analogous results hold for a+b.)
Proposition (25) shows that a second-case solution of Fermat's equation where p divides c implies ap-1≡1(mod p2) and bp-1≡1(mod p2) (avoiding the constraint that p not divide c in Furtwängler's first theorem).
Furtwängler's Theorems and Hasse's Reciprocity Law
This section requires some familiarity with algebraic number theory. In the following, the brackets "{}" denote the pth power residue symbol. Hasse used one of his reciprocity laws to give a more systematic proof of Furtwängler's theorems. Hasse's reciprocity law is; {β/α}{α/β}=ζTr(η) where η=((β-1)/p)((α-1)/λ) for all α,β in Q(ζ) with g.c.d.(α, β)=1, α≡1(mod λ), β≡1(mod p), and where Tr denotes the trace from Q(ζ) to Q. Setting α to (a+ζb)/(a+b) and β to qp-1 where q divides b gives {qp-1/[(a+ζb)/(a+b)]}=ζ-u(p-1)b/(a+b) where u=(qp-1-1)/p since α=1-bλ/(a+b)≡1(mod λ) and α≡1(mod q). If ap+bp=cp, p does not divide c, then α is the pth power of an ideal and hence {β/α}=1 for all β in Q(ζ) that are prime to α. Then if p does not divide b, p must divide u.The question of whether a larger power of p divides (qp-1-1)/p than a+b can be avoided by considering the reciprocal of α. ((a+ζ2b)/(a+ζb))((a+ζ3b)/(a+ζ2b))···((a+ζpb)/(a+ζp-1b))=(a+b)/(a+ζb). Let d(m,n) denote m "choose" n (a binomial coefficient). Let f1 denote d(1,0)ap-2-d(2,1)ap-3b+d(3,2)ap-4b2-...-d(p-1,p-2)bp-2, f2 denote d(2,0)ap-3-d(3,1)ap-4b+d(4,2)ap-5b2-...+d(p-1,p-3)bp-3, f3 denote d(3,0)ap-4-d(4,1)ap-5b+d(5,2)ap-6b2-...-d(p-1,p-4)bp-4, ..., and fp-1 denote d(p-1,0). Collecting terms in the product (1-bλζ/(a+ζb))(1-bλζ2/(a+ζ2b))···(1-bλζp-1/(a+ζp-1b)) gives (a+b)/(a+ζb)=1+(bλf1+b2λ2f2+...+bp-1λp-1ff-1)/((ap+bp)/(a+b)). Tr(1)=p-1 and Tr(λk)=p. Set α to (a+b)/(a+ζb). Substituting for Tr(1), Tr(λ), Tr(λ2), ..., Tr(λp-2) and collecting terms gives Tr((α-1)/λ)=b((p-1)ap-2-(p-2)ap-3b+(p-3)ap-4b2-...-bp-2)/((ap+bp)/(a+b)). Setting β to qp-1 where q divides b gives {qp-1/[(a+b)/(a+ζb)]}=ζuv where v=b((p-1)ap-2-(p-2)ap-3b+(p-3)ap-4b2-...-bp-2)/((ap+bp)/(a+b)). If p divides a+b, then p also divides (ap+bp)/(a+b). (p-1)ap-2-(p-2)ap-3b+(p-3)ap-4b2-...-bp-2 is congruent to -(a+b)p-2 modulo p, therefore if p divides a+b, then 1/p does not divide v. By Proposition (25), if ap+bp=cp, p divides c, then bp-1≡1(mod p2) and hence {bp-1/[(a+b)/(a+ζb)]}=1.
p=(1-ζ)(1-ζ2)(1-ζ3)···(1-ζp-1) and the ideals [1-ζ], [1-ζ2], [1-ζ3], ..., [1-ζp-1] are equal. If p divides a+b and (ap+bp)/(a+b)/p is a pth power, then ((a+ζ2b)/(1-ζ2))/((a+ζb)/(1-ζ)), ((a+ζ3b)/(1-ζ3))/((a+ζ2b)/(1-ζ2)), ((a+ζ4b)/(1-ζ4)/((a+ζ3b)/(1-ζ3)), ..., ((a+ζp-1b)/(1-ζp-1))/((a+ζp-2b)/(1-ζp-2)) are pth power of fractional ideals and hence (a+ζ2b)/(a+ζb), (a+ζ3b)/(a+ζ2b), (a+ζ4b)/(a+ζ3b), ..., (a+ζp-1b)/(a+ζp-2b) are pth powers of fractional ideals. Furthermore, (a+ζi+1b)/(a+ζib)=1-bλζi/(a+ζib)=αi≡1(mod λ), i=1, 2, 3, ..., p-2, and hence {β/αi}=1 for all β in Q(ζ) that are prime to αi. Tr(-bζ/(a+ζb))+Tr(-bζ2/(a+ζ2b))+Tr(-bζ3/(a+ζ3b))+...+Tr(-bζp-1/(a+ζp-1b))=b(ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2)/((ap+bp)/(a+b)), therefore if p divides a+b and (ap+bp)/(a+b)/p is a pth power, then {qp-1/[(a+b)/(a+ζp-1b)]}=ζuw where w=b(ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2)/((ap+bp)/(a+b)). ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2 is congruent to (a+b)p-2 modulo p, therefore 1/p does not divide w. By Proposition (25), if ap+bp=cp, p divides c, then bp-1≡1(mod p2) and hence {bp-1/[(a+b)/(a+ζp-1b)]}=1. The following proposition is based on data collected for p=3;
(26) If 2 divides b, p divides a+b, and (ap+bp)/(a+b)/p is a pth power, then p3 divides (p-1)ap-2-(p-2)ap-3b+(p-3)ap-4b2-...-bp-2 and p2 does not divide ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2. If 2p divides a+b and (ap+bp)/(a+b)/p is a pth power, then p2 does not divide (p-1)ap-2-(p-2)ap-3b+(p-3)ap-4b2-...-bp-2 or ap-2-2ap-3b+3ap-4b2-...-(p-1)bp-2.
By Proposition (5), if 2p-1≠1(mod p2), 2 divides b, p divides a+b, and (ap+bp)/(a+b)/p is a pth power, then bp-1≠1(mod p2). Proposition (26) then implies that if 2p-1≠1(mod p2), 2 divides b, p divides a+b, and (ap+bp)/(a+b)/p is a pth power, then {bp-1/[(a+ζb)/(a+b)]}≠1, {bp-1/[(a+b)/(a+ζb)]}=1, and {bp-1/[(a+b)/(a+ζp-1b)]}≠1.
Vandiver's Theorem
In 1919, Vandiver proved that if ap+bp=cp, p divides c, then p3 divides c, ap-1≡1(mod p3), and bp-1≡1(mod p3). When a is odd, Vandiver's theorem gives a necessary condition for a factor of a to be a pth power w.r.t. (ap+bp)/(a+b) (based on data collected for p=3). Analogous results hold for b, a-b, and a+b. The following proposition is based on data collected for p=3;(27) If [(ap+bp)/(a+b)]=Tp and T=Uk where U is a prime and p does not divide k, 2p divides a, b, a-b, or a+b, 2 does not divide a, q divides a, and qp-1≡1(mod p3), then q is a pth power w.r.t. (ap+bp)/(a+b). If [(ap+bp)/(a+b)] is a pth power, 2p divides a, b, a-b, or a+b, 2 does not divide a, and q divides a, then q is a pth power w.r.t. (ap+bp)/(a+b) only if qp-1≡1(mod p3). (Analogous results hold for b, a-b, and a+b.) If [(ap+bp)/(a+b)]=Tp and T=Uk where U is a prime and p does not divide k, 2 divides a, p does not divide a, q divides a, and every prime factor of q is a pth power modulo p2, then q is a pth power w.r.t. (ap+bp)/(a+b). (Analogous results hold for b.) If [(ap+bp)/(a+b)]=Tp and T=Uk where U is a prime and p does not divide k, p divides a+b, 2 does not divide a+b, q divides a+b, and qp-1≡1(mod p3), then q is a pth power w.r.t. (ap+bp)/(a+b). If [(ap+bp)/(a+b)] is a pth power, p divides a+b, 2 does not divide a+b, q divides a+b, and p does not divide q, then q is a pth power w.r.t. (ap+bp)/(a+b) only if qp-1≡1(mod p3).
Euler's Theorem and "Split" 2 and p
Euler proved that every prime of the form 6k+1 can be represented by x2+3y2. Let T be a natural number and x, y, and z be integers. If p=3, every prime factor of T is of the form 6k+1, and T has n such distinct prime factors, then Tp or pTp has exactly pn representations of the form (ap+bp)/(a+b). All representations of pTp must be of the same type, that is, if (ap+bp)/(a+b) is one representation, then p divides a+b, and if ((a')p+(b')p)/(a'+b') is another representation, then p must divide a'+b'. Representations of Tp can be of different types, that is, p can divide a, b, or a-b. Suppose p=3, (ap+bp)/(a+b) is a representation of pTp, and 2 and p are common factors of a+b. When p=3 and x+y=z, (xp-yp)/(x-y)=(zp+yp)/(z+y)=(zp+xp)/(z+x), so 2 must divide b' where b'=a-b and a'=a for the representation ((a')p+(b')p)/(a'+b') of pTp (and p must divide a'+b') and 2 must divide a'' where a''=a-b and b''=-b for the representation ((a'')p+(b'')p)/(a''+b'') of pTp (and p must divide a''+b'').A Generalization of Vandiver's Theorem
There is some evidence that if there is one representation [(ap+bp)/(a+b)] of Tp for p>3, there must be other representations. Note that if ap+bp=cp, p divides c, and every prime factor of (cp-bp)/(c-b) and (cp-ap)/(c-a) is of the form p2k+1, then p3 divides c by Barlow's formulas (since ap+bp+a+b-2c=rp(Rp-1)+sp(Sp-1)). Vandiver's theorem suggests that no prime factor of (cp-bp)/(c-b) or (cp-ap)/(c-a) can just be of the form pk+1. Vandiver's theorem can be reformulated so that it is applicable to the problem of determining if [(ap+bp)/(a+b)] can be a pth power. The following proposition is based on data collected for p=3;(28) If [(ap+bp)/(a+b)]=Tp where every prime factor of T is of the form p2k+1, p3 divides a, b, or a-b or p4 divides a+b, and 2 does not divide a, then ap-1≡1(mod p3). If [(ap+bp)/(a+b)]=Tp where T has only one distinct prime factor, this prime factor is of the form p2k+1, p3 divides a', b', or a'-b' or p4 divides a'+b' for some representation [((a')p+(b')p)/(a'+b')] of Tp, and 2 does not divide a, then ap-1≡1(mod p3). (Analogous results hold for b and a-b.) If [(ap+bp)/(a+b)]=Tp where every prime factor of T is of the form p2k+1, p3 divides a, b, or a-b or p4 divides a+b, and 2 does not divide a+b, then [(a+b)/p]p-1≡1(mod p3) if p divides a+b, or (a+b)p-1≡1(mod p3) if p does not divide a+b. If [(ap+bp)/(a+b)]=Tp where T has only one distinct prime factor, this prime factor is of the form p2k+1, p3 divides a', b', or a'-b' or p4 divides a'+b' for some representation [((a')p+(b')p)/(a'+b')] of Tp, and 2 does not divide a+b, then [(a+b)/p]p-1≡1(mod p3) if p divides a+b, or (a+b)p-1≡1(mod p3) if p does not divide a+b. If [(ap+bp)/(a+b)]=Tp where every prime factor of T is of the form p2k+1, 2 divides a, and p does not divide a, then (a/2)p-1≡1(mod p3). (Analogous results hold for b.) If [(ap+bp)/(a+b)]=Tp where every prime factor of T is of the form p2k+1, 2 does not divide a, and p divides a+b, then ap-1≡1(mod p3). (Analogous results hold for b and a-b.)
Congruence Properties of Prime Factors of [(ap+bp)/(a+b)] when [(ap+bp)/(a+b)] is a Pth Power
That the reformulation of Vandiver's theorem depends on different representations [(ap+bp)/(a+b)] of Tp is some indication that different representations must exist for p>3 (if there are any representations). Whether p is a pth power w.r.t. (ap+bp)/(a+b) is of importance to Vandiver's theorem. The following propositions are based on data collected for p=3;(29) If [(ap+bp)/(a+b)]=Tp and T=Uk where U is a prime and p does not divide k, then p is a pth power w.r.t. (ap+bp)/(a+b) if and only if p3 divides a', b', or a'-b' or p4 divides a'+b' for some representation [((a')p+(b')p)/(a'+b')] of Tp. If [(ap+bp)/(a+b)]=Tp and T=Uk where U is a prime and p divides k, then p is not a pth power w.r.t. (ap+bp)/(a+b).
(30) If [(ap+bp)/(a+b)] is a pth power, then pp-1a is a pth power w.r.t. (ap+bp)/(a+b) if 2p divides a, pp-1b is a pth power w.r.t. (ap+bp)/(a+b) if 2p divides b, pp-1(a-b) and a+b are pth powers w.r.t. (ap+bp)/(a+b) if 2p divides a-b, or a-b and p(a+b) are pth powers w.r.t. (ap+bp)/(a+b) if 2p divides a+b.
Note that ap+bp=cp, p divides c, implies that p(a+b) is a pth power w.r.t. (ap+bp)/(a+b) and also that 2p divides a+b (if Proposition (5) is accepted). There is no apparent reason (other than the "difficult proof" hypothesis) why p(a+b) should be a pth power w.r.t. (ap+bp)/(a+b) when [(ap+bp)/(a+b)] is a pth power and 2p divides a+b.
(31) If [(ap+bp)/(a+b)] is a pth power, then a is a pth power w.r.t. (ap+bp)/(a+b) if 2 divides a and p does not divide a, or b is a pth power w.r.t. (ap+bp)/(a+b) if 2 divides b and p does not divide b.
(32) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap+bp)/(a+b)] of the form p2k+1, and p is not a pth power modulo f, then (1) pp-1a, pp-1b, pp-1(a-b), and a+b are pth powers modulo f if p divides a, b, or a-b, or (2) a, b, a-b, and p(a+b) are pth powers modulo f if p divides a+b. If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap+bp)/(a+b)] of the form p2k+1, and p is a pth power modulo f, then a, b, a-b, and a+b are pth powers modulo f.
Note that ap+bp=cp, p divides c, implies every prime factor of [(ap+bp)/(a+b)] is of the form p2k+1 and also that a and b are pth powers w.r.t. (ap+bp)/(a+b). There is no apparent reason (other than the "difficult proof" hypothesis) why a and b should be pth powers modulo f when [(ap+bp)/(a+b)] is a pth power, p divides a+b, and f is a prime factor of [(ap+bp)/(a+b)] of the form p2k+1.
(33) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap+bp)/(a+b)] not of the form p2k+1, and p is not a pth power modulo f, then (1) pp-1a, b, p(a-b), and pp-1(a+b), or pp-1a, pb, a-b, and p(a+b) are pth powers modulo f if 2p divides a, or (2) a, pp-1b, p(a-b), and pp-1(a+b), or pa, pp-1b, a-b, and p(a+b) are pth powers modulo f if 2p divides b, or (3) a, pb, pp-1(a-b), and a+b, or pa, b, pp-1(a-b), and a+b are pth powers modulo f if 2p divides a-b, or (4) pa, pp-1b, a-b, and p(a+b), or pp-1a, pb, a-b, and p(a+b) are pth powers modulo f if 2p divides a+b.
(34) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap+bp)/(a+b)] not of the form p2k+1, and p is not a pth power modulo f, then (1) a, pb, pp-1(a-b), and a+b, or a, pp-1b, p(a-b), and pp-1(a+b) are pth powers modulo f if 2 divides a and p does not divide a, or (2) pa, b, pp-1(a-b), and a+b, or pp-1a, b, p(a-b), and pp-1(a+b) are pth powers modulo f if 2 divides b and p does not divide b.
Note that since Propositions (30), (32), (33), and (34) are based solely on data collected for p=3, their form is ambiguous in that the p exponents might be 2 instead of p-1. Congruence properties of the prime factors of [(ap-bp)/(a-b)] when [(ap+bp)/(a+b)] is a pth power appear to determine the form of Propositions (30) through (34). (Propositions (30), (31), (32), (33), and (34) can be transformed into Propositions (11), (12), (8), (14), and (9) respectively by multiplying the a, b, a-b, and a+b terms by p and switching the a+b and a-b terms [and of course the modulo bases]. This is just an attempt to find a simple relationship between the congruence properties of the prime factors of [(ap+bp)/(a+b)] and [(ap-bp)/(a-b)] when [(ap+bp)/(a+b)] is a pth power and has no apparent logical basis.)
(35) If [(ap+bp)/(a+b)] is a pth power, f is a prime factor of [(ap+bp)/(a+b)] not of the form p2k+1, and p is a pth power modulo f, then (1) a (and not b, a-b, or a+b) is a pth power modulo f if 2 divides a, or (2) b (and not a, a-b, or a+b) is a pth power modulo f if 2 divides b, or (3) a-b and a+b (and not a or b) are pth powers modulo f if 2 divides a-b or a+b.
As shown previously, ap+bp=cp, p divides c, implies c, pb, and c-b are pth powers w.r.t. (cp-bp)/(c-b), c, pa, and c-a are pth powers w.r.t. (cp-ap)/(c-a), and 2 and p are common factors of c. Then by Propositions (33) and (35), every prime factor of (cp-bp)/(c-b) and (cp-ap)/(c-a) must be of the form p2k+1. (Substituting c for a and -b for b in Proposition (33) gives pp-1c, -b, p(c+b), and pp-1(c-b), or pp-1c, -pb, c+b, and p(c-b) are pth powers modulo f [a prime factor of (cp-bp)/(c-b)] if 2p divides c and p is not a pth power modulo f [a contradiction]. Substituting c for a and -b for b in Proposition (35) gives c [and not c-b] is a pth power modulo f if 2p divides c and p is a pth power modulo f [a contradiction]. Similar results follow by substituting c for a and -a for b in Propositions (33) and (35).) Furthermore, by Proposition (32), p must be a pth power w.r.t. (cp-bp)/(c-b) and (cp-ap)/(c-a). As shown previously, ap+bp=cp, p divides c, implies a, b, and p(a+b) are pth powers w.r.t. (ap+bp)/(a+b), and every prime factor of [(ap+bp)/(a+b)] is of the form p2k+1. This gives the following proposition;
(36) If ap+bp=cp where p divides c, then every prime factor of (cp-bp)/(c-b) is of the form p2k+1 and c, b, c-b, c+b, and p are pth powers w.r.t. (cp-bp)/(c-b). (Analogous results hold for (cp-ap)/(c-a).) If ap+bp=cp where p divides c, then every prime factor of [(ap+bp)/(a+b)] is of the form p2k+1 and a, b, a-b, and p(a+b) are pth powers w.r.t. (ap+bp)/(a+b).
Further evidence for the above proposition is given by the following proposition (based on data collected for p=3, 5, 7, and 11);
(37) If every prime factor of [(ap+bp)/(a+b)] is of the form p2k+1, and p2 divides a, b, a-b, or a+b, then ap-1≡1(mod p2) if p does not divide a, bp-1≡1(mod p2) if p does not divide b, and (a-b)p-1≡1(mod p2) and (a+b)p-1≡1(mod p2) if p does not divide a-b or a+b.
Substituting c for a and -b for b in Proposition (14) gives p2(c-b) is a pth power modulo f (a prime factor of (cp+bp)/(c+b) not of the form p2k+1) if p and 2p are not pth powers modulo f, or p(c-b) is a pth power modulo f if p is not a pth power modulo f and 2p is a pth power modulo f. Then if c-b is a pth power, p must be a pth power modulo f and hence by Proposition (7), 2 cannot be a pth power modulo f (otherwise, 2p would be a pth power modulo f). As shown previously, ap+bp=cp implies 2 is a pth power w.r.t. (cp+bp)/(c+b). This gives the following proposition;
(38) If ap+bp=cp where p divides c, then every prime factor of (cp+bp)/(c+b) is of the form p2k+1 and c, b, p(c+b) and c-b are pth powers w.r.t. (cp+bp)/(c+b). (Analogous results follow for (cp+ap)/(c+a).)
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Bericht über neuere Undersuchungen und Probleme aus der Theorie der algebraischen Zahlkörper. Teil I: Klassenkörperietheorie. Teil Ia: Beweise zu Tiel I; Teil II: Reziprozitätsgesetz. 3. Aufl.. Wurzburg Wien: Physika-Verlag. 204 pp. Jbuch 52-150; 53, 143; 46-165
