Every integer,i.e. whole number, has a unique sum, ∫. Its sum is the addition of all integers from zero to that specified. For example, the sum of 1 is 1 (0+1), for ∫2 it is 3 (0+1+2), ∫3 is 6 (0+1+2+3) and so on. Mathematically this adding process may be carried out more simply for an integer i by utilising the function 1/2(i^2 + i). Since i, not to be confused with the square root of minus one (√-1), indicates any whole number, the summing function may be written equally well as 1/2[(in)^2 + (in)]. The two formulations dispense with any importance attaching to the exponent n.

When Fermat proposed in 1637 that there were no solutions to cn = an + bn if n was greater than 2 (the last of his theorems to be proven and so abbreviated to FLT) he may well have used the sum of the integer as the basis for his claim knowing that n no longer affected the outcome. An equation requires that one integer, and its unique sum, is the mutual result between sides. Because of the combination of exponents in the ?summing? process two sums of integers cannot equate with a third whilst simultaneously allowing the equation of the numbers naturally. Consequently the sum of an integer may not be divided into two so that when added together derivable integers equate with the original whole number. Otherwise an impossibility arises and the integers of a Pythagorean triple equate naturally with an absurd result : -

∫i?s: 1/2[(c)^2 + (c) ] = 1/2[{(a)^2+(b)^2}+{(a)+(b)}]

both: [(c)^2 - {(a)^2+(b)^2}] = 0

and: [(c) - {(a) +(b) }] = 0

so: [(c)^2 ] = c(a +b)

thus: [ {(a)^2+(b)^2}] = c(a +b) and c = a or b !

The solution requires that the expression ?cn = (a + b)n? substitutes ?cn = an + bn? in the original format. At n > 2 the original ?equation? only works when a = b and each is 1/2c - a truly ?remarkable? metamorphosis.

Professor Wiles achieved proof of the Taniyama-Shimura conjecture (TSC) in 1995 and was thereby accredited with having solved FLT by way of Gerhard Frey?s conjecture of 1984. This supposed a ?solution? CN = AN + BN from which a derived elliptic y^2 = x^3 + (AN - BN)x^2 - ANBN might be obtained. The elliptic was known to be extraordinary and, seemingly having no modular equivalent, contravened TSC. If TSC was proven correct there could be no such elliptic, a supposed ?solution? could not exist and there could be no counter example to Fermat?s statement. As stated ∫CN cannot equate with ∫AN + ∫BN so the conjecture was false ab initio. The ?link? between TSC and FLT is now expunged and, regrettably, accreditation of FLT?s solution to Andrew Wiles is no longer valid despite his undoubted proof of TSC.

Notes :

i Computer programs, necessarily limited in scope, appear to defend the logic to infinity.

ii The logic vindicates Fermat (vide Pythagoras) despite his sceptics.

iii The ∫cn = cn [1/2(an + bn +1)+1/2(cn - (an + bn))] whatever integers are chosen for a, b, c, and n.

iv FLT poses further profound questions for accepted number theory wisdom.

v Beal?s ?xm + yn = zr?, in which x,y and z are co-prime and exponents m,n and r > 2, cannot exist.

THE PROOF

1. Let c,a and b be positive integers each having a unique sum ?∫?

1.1 Given ∫i is the sum of all integers from 1 to that selected i.e ∫1=1, ∫2=3, ∫3=6 et seq.

1.2 ∫i, ∫in = 1/2[i^2+i ],1/2[(in)^2+in] :proven by induction

1.3 ∫ , ∫ = 1/2[i +1 ],1/2[(in) +1 ] :summing operator

1.4 i = in :?∫? operates on any integer

2. To Prove cn ≠ an+bn when n > 2 :Diophantus? Arithmetika

3. Proof of Fermat?s annotation to Problem 8 - ?FLT?

4. when ∫cn =∫an+∫bn :as in ∫8{36} =∫5{15}+∫6{21}

5. 1/2[(cn)^2+cn]≠1/2[(an)^2+(bn)^2+(an+bn)]:i^2≠i (excl.0,1)

6. all cases ∫cn = cn[1/2(an+bn+1)+1/2(cn-(an+bn))]:whatever a,b,c, and n

7. and ∫cn = cn[1/2(an+bn+1)] :if cn = an+bn

8. giving ∫cn = cn[1/2(cn +1)] :(1.3)

9. since cn may = an+bn :when n = 1,2

10. then ∫cn =∫(an+bn) :only equality format(1)so

11. 1/2[(cn)^2+cn]=1/2[cn((a+b)n+1)] :to retain ∫in for n > 2(7)

12. and (an+bn)=(a+b)n :n constrained to 1

13. therefore cn ≠ an+bn :if n > 2



QED

NB:

The Beal Conjecture is confirmed in that no integers satisfy the equation xm+yn = zr in which x,y and z are co-prime and exponents m,n and r > 2. Lines 4 to 13 prohibit equality for ∫(xm+yn)=∫(zr) since 1/2[(xm+yn)^2+(xm+yn)]≠ 1/2[(zr)^2+zr] if exponents > 2.