51st Ukrainian National Mathematical Olympiad, 4th Round

To the contrary, assume that $ac+bd$ is prime. Then $ac+bd\mid ab+cd\Rightarrow ac+bd\mid ac+bd+ab+cd=(a+d)(b+c)$, and so $ac+bd\mid a+d$ or $ac+bd\mid b+c$. But this is impossible because for distinct $a,b,c,d$ we have that $ac+bd>a+d$ and $ac+bd>b+c$. This contradiction completes the proof.