Problem: Let $a$, $b$, $c$, $d$, $e$ be distinct positive integers
such that $a^4+b^4=c^4+d^4=e^5$.
Show that $ac+bd$ is a composite number.

Solution: Assume to the contrary that $p = ac+bd$, so that
\begin{align*}
  ac &\equiv -bd \pmod p \\
  \implies a^4c^4 &\equiv b^4d^4 \pmod p \\
  \implies a^4 (e^5 - d^4) &\equiv (e^5 - a^4) d^4 \pmod p \\
  \implies a^4 e^5 &\equiv d^4 e^5 \pmod p \\
  \implies e^5(a^4-d^4) &\equiv 0 \pmod p
\end{align*}
and hence \[ p \mid e^5(a-d)(a+d)(a^2+d^2). \]

\begin{claim*}
  We should have $p > e$.
\end{claim*}
\begin{proof}
  We have $e^5 = a^4 + b^4 \le a^5 + b^5 < (ac+bd)^5 = p^5$.
\end{proof}
Thus the above equation implies $p \le \max(a-d, a+d, a^2+d^2) = a^2+d^2$.
Similarly, $p \le b^2+c^2$.
So \[ ac+bd = p \le \min \left\{ a^2+d^2, b^2+c^2 \right\} \]
or by subtraction
\[ 0 \le \min \left\{ a(a-c) + d(d-b),
  b(b-d) + c(c-a) \right\}. \]
But since $a^4+b^4 = c^4+d^4$ the numbers $a-c$ and $d-b$
should have the same sign, and so this is an obvious contradiction.