Let $a,b,c,d\in N$ and in strictly increasing order such that $b^2-bd-d^2=a^2-ac-c^2$. Prove that $ab+cd$ is not a prime.

Question

Let $a,b,c,d\in N$ and in strictly increasing order such that $$b^2-bd-d^2=a^2-ac-c^2$$ Prove that $ab+cd$ is not a prime.

Approach: $c^2-d^2-bd=a^2-b^2-ac$

$\implies c^2-d^2-bd=a^2-b^2-ac+bc-bc$

$\implies c^2-d^2-bd=(a+b)(a-b)-c(a-b)-bc$

$\implies c^2-d^2-bd+bc=(a+b)(a-b)-c(a-b)$

$\implies c^2-d^2-bd+bc=(a-b)(a+b-c)$

$\implies (c+d)(c-d)+b(c-d)=(a-b)(a+b-c)$

$\implies (c+d+b)(c-d)=(a-b)(a+b-c)$

How to processed further.?

This question is similar to Prove ${a^2+ac-c^2=b^2+bd-d^2}$ and $a > b > c > d \implies ab + cd$ is not prime But he used factoring lemma and provided link but that link is not working.

Can you show me how to do that?

Is there other method too to solve this problem?

Answer 1

This works if $a>b>c>d>0$.

Write $p= ab+cd$ and assume $p$ is prime. Then plug in $a = (p-cd)/b$ and we get: $$\boxed{p(ab-cd-bc) = (b^2+c^2)(b^2-bd-d^2)}$$ and if we plug $c= (p-ab)/d$ we get $$\boxed{p(ab-cd-ad) = (a^2+d^2)(b^2-bd-d^2)}$$

1. If $p\mid b^2-bd-d^2$

   • If $b^2-bd-d^2 \geq 0$ then $ab+cd \leq b^2 -bd -d^2<b^2<ab$ a contradiction.
   • If $b^2-bd-d^2 < 0$ then $ab+cd \leq d^2+bd -b^2 <d^2+bd <ab +cd$ a contradiction.

2. If $p\nmid b^2-bd-d^2$, then $p\mid a^2+d^2$ and $p\mid b^2+c^2$ so $$p^2\mid (a^2+d^2)(b^2+c^2) = (ab+cd)^2 +(ac-bd)^2$$ and thus $$p^2\mid (ac-bd)^2\implies p\mid ac-bd \implies ab+cd\leq ac-bd< ac<ab$$ A contradiction again. So $p$ can not be prime.