Selection tests for the Gulf Mathematical Olympiad 2013

Let $p$ be a prime divisor of $n^{55}-n$ for all integer $n$. Whenever $n$ is not divisible with $p$, we have $$n^{54}\equiv 1\mathrm{mod}p$$ In this case, the order of $n$ modulo $p$ divides $54$. But there exists an integer $n$ of order $p-1$ modulo $p$. We deduce that $p-1$ divides $54$. But the only primes $p$ such that $p-1$ divides $54$ are $p=2,3,7$ and $19$. Conversely, all these four primes $2,3,7$ and $19$ divide $n^{55}-n$ for all integer $n$ by Fermat's little theorem. Notice that for $p$, $p^{55}-1$ is not divisible by $p^2$ for all prime numbers $p$ since $p^{54}-1$ is relatively prime with $p$. Therefore, the greatest integer $k$ which divides $n^{55}-n$ for all integer $n$ is $2\times 3\times 7\times 19=798$.