Let $a, b, c, d$ be positive integers satisfying $ab=cd$. Prove that $a+b+c+d$ ݀ is composite.

Question

I started with an even odd arguement, where I tried to show that no matter which numbers you pick, the sum would always be even. However, this doesn't work for the arrangment of 3 evens and one odd. Where do I go from here?

Answer 1

Write $p=a+b+c+d$ and say $p$ is prime. Then we have $$ab=c(p-a-b-c)$$ so $$(a+c)(b+c) = cp$$ which means that $$p\mid a+c\;\;\;\;{\rm or}\;\;\;\;p\mid b+c$$ in 1st case we get $p=a+b+c+d\leq a+c$ a contradiction. The same contradiction we get in the second case. So $p$ must be composite.