, 2013

Let $p$ be a prime divisor of $n^{55} - n$ for all integer $n$. Whenever $n$ is not divisible with $p$, we have
$$n^{54} \equiv 1 \quad \bmod p$$
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 $2 \times 3 \times 7 \times 19 = 798$.