Thailand Mathematical Olympiad

We first show that $3 \mid (p^4 - 1)$. Since $p > 3$ is a prime, we have $3 \nmid p$, and so $p \equiv \pm 1 \pmod{3}$. Thus, $p^4 \equiv 1 \pmod{3}$, as required.

We next claim that $2^4 \mid (p^4 - 1)$. Since $p$ is odd, we have $p \equiv \pm 1, \pm 3, \pm 5, \pm 7 \pmod{16}$. If $p \equiv \pm 1 \pmod{16}$, then, clearly, $p^4 \equiv 1 \pmod{16}$. The remaining possibilities are similarly checked, and the claim follows.

Since $\gcd(2^4, 3) = 1$, we have $2^4 \times 3 \mid (p^4 - 1)$. Let $d \in \mathbb{N}$ be a divisor of $p^4 - 1$ for all primes $p > 3$. Thus, $d \mid (5^4 - 1)$ and $d \mid (7^4 - 1)$ yielding $d \mid \gcd(5^4 - 1, 7^4 - 1) = 2^4 \times 3$.