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,
$$x^{p-1} \equiv y^{p-1} \equiv 1 \quad \bmod p$$
But $p-1$ divides $1440$ for $p = 3, 11, 61$. We deduce that
$$x^{1440} \equiv y^{1440} \equiv 1 \quad \bmod p .$$
Hence
$$x^{7} + y^{7} \equiv x^{7} y^{1440} + y^{7} x^{1440} \equiv x^{7} y^{7} (y^{1433} + x^{1433}) \equiv 0 \quad \bmod p .$$
This proves that $p$ divides $x^{7} + y^{7}$.