mathschallenge.net

Problem

Given that $p$ is prime, when is $4^p + p^4$ prime?

Solution

For any prime, $p \gt 5$, $p \equiv 1, 3, 7, 9 \mod 10$.

If $p \equiv 1, 9 \mod 10$, then $p^2 \equiv 1 \mod 10$, and $p^4 \equiv 1 \mod 10$.

If $p \equiv 3, 7 \mod 10$, then $p^2 \equiv -1 \mod 10$, and $p^4 \equiv 1 \mod 10$.

Therefore, for $p \gt 5$, $p^4 \equiv 1 \mod 5$.

We note that $4^1 \equiv 4 \mod 5$. Now if $4^m \equiv 4 \mod 5$, then multiplying both sides by $4^2$ gives $4^{m+2} \equiv 4 \mod 5$. That is, if $n$ is odd then $4^n \equiv 4 \mod 5$.

Hence, for $p \gt 5$, $4^p + p^4 \equiv 0 \mod 5$, and cannot be prime.

When $p = 2$, $4^p + p^4 = 32$.
When $p = 3$, $4^p + p^4 = 145$.
When $p = 5$, $4^p + p^4 = 1649 = 17 \times 97$.

Hence the expression $4^p + p^4$ is never prime.

See Prime Exponent And Square Sum.