SAUDI ARABIAN MATHEMATICAL COMPETITIONS

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