Saudi Arabian Mathematical Competitions

By contrary, assume that $p = a b + c d$ is a prime. In particular $(b, c) = (b, d) = 1$, and $a b \equiv -c d \pmod{p}$. We get
$$\begin{aligned} 0 & = b^{2} (b^{2} + b d - d^{2}) - b^{2} (a^{2} + a c - c^{2}) \\ & = b^{2} (b^{2} + b d - d^{2}) - (a b)^{2} - (a b)(b c) + b^{2} c^{2} \\ & \equiv (b^{2} + c^{2})(b^{2} + b d - d^{2}) \pmod{p} \end{aligned}$$
Therefore, one of the numbers $b^{2} + c^{2}$ and $b^{2} + b d - d^{2}$ is divisible by $p$. If $p \mid b^{2} + c^{2}$, then $b^{2} + c^{2} = p$ since $0 < b^{2} + c^{2} < 2(a b + c d) = 2p$. This shows $a b + c d = b^{2} + c^{2}$ and then $b \mid c(c - d)$. But $b, c$ are coprime, we get $b \mid c - d$ which is impossible.
If $p \mid b^{2} + b d - d^{2}$, then $b^{2} + b d - d^{2} = p$ since $0 < b^{2} + b d - d^{2} < 2 b^{2} < 2 a b < 2p$. This shows that $a b + c d = b^{2} + b d - d^{2}$ and then
$$b(a - b) = b d - c d - d^{2} = d(b - c - d)$$
In particular, $b - c - d > 0$ and $b \mid b - c - d$ (since $b$, $d$ are coprime) which is impossible, again.
Hence, $a b + c d$ must be a composite number.