# x^3+y^3=z^3

• January 3rd 2007, 11:14 AM
Jenny20
x^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.
• January 3rd 2007, 11:22 AM
ThePerfectHacker
Quote:

Originally Posted by Jenny20
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.

Work modulo 7.
$(7k)^3=7$
$(7k+1)^3=7k+1$
$(7k+2)^3=7k+1$
$(7k+3)^3=7k+6$
$(7k+4)^3=7k+6$
$(7k+5)^3=7k+6$
$(7k+6)^3=7k+6$

Possibilities,
$7k,7k+1,7k+6$

We assume $\gcd(x,y,z)=1$.
In that case not all can be multiple of sevens.

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