Determine the largest positive integer $n$ that divides $p^{6}-1$ for all primes $p>7$.

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Note that

$$
p^{6}-1=(p-1)(p+1)\left(p^{2}-p+1\right)\left(p^{2}+p+1\right)
$$

For $p=11$ we have

$$
p^{6}-1=1771560=2^{3} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 19 \cdot 37
$$

For $p=13$ we have

$$
p^{6}-1=2^{3} \cdot 3^{2} \cdot 7 \cdot 61 \cdot 157
$$

From the last two calculations we find evidence to try showing that $p^{6}-1$ is divisible by $2^{3} \cdot 3^{2} \cdot 7=504$ and this would be the largest positive integer that divides $p^{6}-1$ for all primes greater than 7.

By Fermat's theorem, $7 \mid p^{6}-1$.

Next, since $p$ is odd, $8 \mid p^{2}-1=(p-1)(p+1)$, hence $8 \mid p^{6}-1$.

It remains to show that $9 \mid p^{6}-1$.

Any prime number $p, p>3$ is 1 or -1 modulo 3.

In the first case both $p-1$ and $p^{2}+p+1$ are divisible by 3, and in the second case, both $p+1$ and $p^{2}-p+1$ are divisible by 3.

Consequently, the required number is indeed 504.

## Alternative solution

Let $q$ be a (positive) prime factor of $n$. Then $q \leq 7$, as $q \nmid q^{6}-1$. Also, $q$ is not 5, as the last digit of $13^{6}-1$ is 8.

Hence, the prime factors of $n$ are among 2, 3, and 7.

Next, from $11^{6}-1=2^{3} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 19 \cdot 37$ it follows that the largest integer $n$ such that $n \mid p^{6}-1$ for all primes $p$ greater than 7 is at most $2^{3} \cdot 3^{2} \cdot 7$, and it remains to prove that 504 divides $p^{6}-1$ for all primes greater than 7.