More generalize problem: Let a, b, c, d be integers with $a>b>c>d>0$. Suppose that $ac+bd =k(b+d+a-c)$ $k \in \mathbb{Z}$. Prove that $ab+cd$ is not prime.

Hmm, that is a nice result. Do you have a proof for it?

Yes, but this result isn't of me. It's form an article form an author in Vietnam (I can't remember his name). This is the solution Assume that $ab+cd$ is prime. We have $ab+cd=(a+d)c+(b-a)a=m(a+d,b-c)$ $m\in \mathbb{N}$ Case $m=1$ We have $ab+cd=(a+d,b-c) < a+d$ (contradict) case $ (a+d,b-c)=1$ Because $b+d+a-c|ac+bd$ so exist $p$ such that $ac+bd=p(b+d+a-c)$ In the other hand, $ac+bd=(a+d)b-(b-c)a=p(b+d+a-c)=p(a+d)-(b-c)p$ So $(p-b)(a+d)=(a+p)(b-c)$ But$ (a+d,b-c)=1$ which follows that $a+p=(a+d)k \rightarrow b-p=k(b-c)$ $\rightarrow a+b=k(a+b+d-c)$ If $k=1$ then $c=d$ (contradict) If $k\ge 2$ then $a+b\ge 2(a+b+d-c) \rightarrow 2c\ge a+b+2d $ (contradict)

Peter wrote: Let $a, b, c, d$ be integers with $a\geq b>c>d\geq 0$. Suppose that $ac+bd=(b+d+a-c)(b+d-a+c)$. Prove that $ab+cd$ is not prime. $(ab+cd)(ab-cd-bc)$ $=\left(b^2-c^2\right)\left(c^2-ca+a^2\right)+c^2\left[ac+bd-(b+d+a-c)(b+d-a+c)\right]$ $=\left(b^2-c^2\right)\left(c^2-ca+a^2\right)$, where $(2a+b-c+2d)(ab-cd-bc)$ $=(b-c)\left[a^2+b^2+c^2+d^2+bc+2ad+(b-c)(a+d)\right]$ $+(b+c)\left[ac+bd-(b+d+a-c)(b+d-a+c)\right]$ $=(b-c)\left[a^2+b^2+c^2+d^2+bc+2ad+(b-c)(a+d)\right]>0$ $\Longrightarrow ab-cd-bc>0$, $(a+b-c+d)\left[\left(b^2-c^2\right)-(ab-cd-bc)\right]$ $=(b-c)(c-d)(d+a-c)-b\left[ac+bd-(b+d+a-c)(b+d-a+c)\right]$ $=(b-c)(c-d)(d+a-c)>0$ $\Longrightarrow b^2-c^2>ab-cd-bc$, $\left(c^2-ca+a^2\right)-(ab-cd-bc)$ $=cd+c^2-c(a-b)+(a-b)^2+b(a-b)>0$ $\Longrightarrow c^2-ca+a^2>ab-cd-bc$.

up to my calculations b|a and c|a is that true?

ISL 2001 N5