Preselection tests for the full-time training

Since $2013=3\times 11\times 61$, we will prove that for $p=3,11,61$, if $p$ divides $x^{1433}+y^{1433}$ then $p$ divides $x^7+y^7$.

Let $p=3,11,61$, and assume that $p$ divides $x^{1433}+y^{1433}$.

If $p$ divides $x$, then it divides $x^7$ and $x^{1433}$. But $p$ divides $x^{1433}+y^{1433}$. Then it divides $y^{1433}$. Since $p$ is a prime number, we deduce that it divides $y$ and $y^7$. Therefore, it divides $x^7+y^7$. In a similar way we prove that if $p$ divides $y$, then it divides $x^7+y^7$.

Assume now that $p$ is relatively prime with $x,y$. Then, using Fermat, $$x^{p-1}\equiv y^{p-1}\equiv 1\mathrm{mod}p$$ But $p-1$ divides $1440$ for $p=3,11,61$. We deduce that $$x^{1440}\equiv y^{1440}\equiv 1\mathrm{mod}p.$$ Hence $$x^7+y^7\equiv x^7{y}^{1440}+y^7{x}^{1440}\equiv x^7{y}^7(y^{1433}+x^{1433})\equiv 0\mathrm{mod}p.$$ This proves that $p$ divides $x^7+y^7$.