TITLE: Prove ${a^2+ac-c^2=b^2+bd-d^2}$ and $a > b > c > d \implies ab + cd$ is not prime
QUESTION [6 upvotes]: Let $a>b>c>d$ be positive integers and suppose that
$${a^2+ac-c^2=b^2+bd-d^2}$$
Prove that $ab+cd$ is not prime? I don't know if this problem is true.
I found that this same problem has also been posted on AOPS.
But I can't prove this problem. Can anyone help me?

REPLY [2 votes]: The key observation is:
$$ (ac-bd) ( a^2 + ac - c^2) =  (ab+cd) ( bc - ad). $$
This can be shown by expanding the LHS as $(ac)(b^2+bd-d^2) - (bd) (a^2 + ac - c^2)$, noticing that $abcd$ cancels out, and factoring.
Proof by contradiction. Suppose $ab+cd$ is prime.

Because $ac-bd > 0, a^2+ac-c^2 > 0, ab+cd > 0,$ hence we can conclude that $bc-ad > 0$.
Since $0 < ac - bd < ab + cd$, so $\gcd(ac-bd, ab+cd) = 1$, hence $ ac-bd \mid bc-ad$.
This implies that $ ac-bd \leq bc-ad$.
However, $(a-b)(c+d) > 0  \Leftrightarrow ac - bd > bc  - ad $.

We have a contradiction, so $ ab+cd$ is not prime.

Notes
I came up with the key observation because I was working on similar problems like

IMO 2001/6,
If $a^2+ab+b^2 = c^2 + cd+d^2$, then $a+b+c+d$ is not prime,   and
If $ a>b>c >d > 0 $ such that $a^2-ab+b^2 = c^2 - cd + d^2 $ then $ ab+cd $ is not prime.

In all of them, there's an approach that similarly finds a nice algebraic expression involving the variables, then argues about primality.