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

• Jan 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.
• Jan 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.
\$\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.