# X^p + p*M + [Y-X] = Y^p

1. Aug 29, 2004

### Russell E. Rierson

A general equation for Y > X :

X^p + p*M + [Y-X] = Y^p

3^3 + 3*12 + 1 = 4^3

3^3 + 3*32 + 2 = 5^3

3^3 + 3*62 + 3 = 6^3

3^3 + 3*104 + 4 = 7^3

3^3 + 3*160 + 5 = 8^3

3^3 + 3*232 + 6 = 9^3

[...]
4^3 + 3*20 + 1 = 5^3

4^3 + 3*50 + 2 = 6^3

4^3 + 3*92 + 3 = 7^3

4^3 + 3*148 + 4 = 8^3

4^3 + 3*220 + 5 = 9^3

[...]

X^p + p*M + N = Y^p

It also works for p = 2,5,7, etc...?

3^2 + 2*3 + 1 = 4^2

3^2 + 2*7 + 2 = 5^2

3^2 + 2*12 + 3 = 6^2

3^2 + 2*18 + 4 = 7^2

[...]

4^5 + 5*420 + 1 = 5^5

4^5 + 5*1350 + 2 = 6^5

4^5 + 5*3156 + 3 = 7^5

[...]

X^p + p*M + [Y - X] = Y^p

2. Aug 29, 2004

### Zurtex

Correct me if I am wrong, but to prove that you need to show that for y > x that p is a factor of $y^p - y - x^p + x$. I don't know how to do that, but perhaps you do.

3. Aug 29, 2004

### matt grime

Don't you? You just pick examples where it's true and don't bother about the obvious fact it's false.

4. Aug 29, 2004

### Zurtex

Really? I've not got something on me where I can easily write a program to find a counter example but running a few million values through Excel I can't find one.

5. Aug 29, 2004

### matt grime

actually, i'm doing a slight disservice there, since fermat's little theorem gives you that it is always true. i got carried away with the fact it was something russell wrote and was probably either trivial or false. had i looked more closely i'd have realized it was the first of those. if he ever actually explained using words what he meant he might be clearer

Last edited: Aug 29, 2004
6. Aug 29, 2004

### Russell E. Rierson

X^p + p*M + [Y-X] = Y^p

Y^p - X^p = p*M + [Y-X]

p*M + [Y-X]

is not a "pth" power for p > 2

7. Aug 29, 2004

### Zurtex

Eh? Matt said Fermat's little theorem not Fermat's last theorem.