h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

That is the question. Why do two cubes not result in another cube?



3^2 = 5^2 - 4^2

5^2 = 13^2 - 12^2

7^2 = 25^2 - 24^2

9^2 = 41^2 - 40^2

11^2 = 61^2 - 60^2

etc...



[x+1]^2 - x^2 = 2x+1

2x+1 gives 1st powers, squares, cubes, 4th powers, ...n-powers.

[x+1]^3 - x^3 = 3x^2 + 3x + 1

= 3x[x+1] + 1

3x is a cube

[x + 1] is a cube

3x[x+1] + 1 is not a cube


Let a^3 = 3x, Let b^3 = [x+1].

integer a > 0 , integer b > 0

[ab]^3 + 1 is not a cube


Q@A = A@Q


[x+y]^2 - x^2 = y[2x+y]

5^2 - 4^2 = 3^2

[x+y]^3 - x^3 = y[3x^2 + 3xy + y^2]

y[2x+y] IS a square

y[3x^2 + 3xy + y^2] is NOT a cube

y[4x^3 + 6yx^2 + 4xy^2 + y^3] is NOT a 4th power

y[5x^4 + 10yx^3 + 10(xy)^2 + 5xy^3 + y^4] is NOT a 5th power

For example if y = 1

3x^2 + 3x + 1 cannot be a cube?

if y = 2

2*[3x^2 + 6x + 4] = cannot be a cube?

if y = 3

3*[3x^2 + 9x + 9] cannot be a cube?


For the simple case x = x and y = 1:

[x+1]^3 - x^3 = 3x^2 + 3x + 1

equals

6*x*[x+1]/2 + 1

6*1+1 = 6+1

6*2+1 = 12+1

6*3+1 = 18+1

6*6+1 = 36+1

etc...

equals

6*N*[N+1]/2 + 1

equals

6*[1+2+3+...+ N] + 1

not a cube...

y[3x^2 + 3xy + y^2]

x = x

y = n

1[3x^2 + 3x + 1]

2[3x^2 + 6x + 4]

3[3x^2 + 9x + 9]

4[3x^2 + 12x + 16]

5[3x^2 + 15x + 25]

6[3x^2 + 18x + 36]

etc...

etc...

etc...

Now let x = 1, with y = n

1[3+3+1] = 2^3 - 1^3

2[3+6+4] = 3^3 - 1^3

3[3+9+9] = 4^3 - 1^3

4[3+12+16] = 5^3 - 1^3

5[3+15+25] = 6^3 - 1^3

6[3+18+36] = 7^3 - 1^3

etc...

etc...

etc...