Question

If x^3+y^3=z^3 has a solution in integers x, y, z, show that one of the three must be a multiple of 7.

How should I solve this question?

Thank you very much.

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- Jan 3rd 2007, 11:14 AMJenny20x^3+y^3=z^3
Question

If x^3+y^3=z^3 has a solution in integers x, y, z, show that one of the three must be a multiple of 7.

How should I solve this question?

Thank you very much. - Jan 3rd 2007, 11:22 AMThePerfectHacker
Work modulo 7.

$\displaystyle (7k)^3=7$

$\displaystyle (7k+1)^3=7k+1$

$\displaystyle (7k+2)^3=7k+1$

$\displaystyle (7k+3)^3=7k+6$

$\displaystyle (7k+4)^3=7k+6$

$\displaystyle (7k+5)^3=7k+6$

$\displaystyle (7k+6)^3=7k+6$

Possibilities,

$\displaystyle 7k,7k+1,7k+6$

We assume $\displaystyle \gcd(x,y,z)=1$.

In that case not all can be multiple of sevens.

The possibilities that do not lead to contradiction,

$\displaystyle (7k)+(7k+1)=(7k+1)$

$\displaystyle (7k)+(7k+6)=(7k+6)$

$\displaystyle (7k+1)+(7k+6)=(7k)$

$\displaystyle (7k+6)+(7k+1)=(7k)$

We see a multiple of seven does exist.