I came up with a solution to a number theory problem. Please check it.
Prove that $a^2 + b^2 + c^2 + d^2$ is never a prime if $ad=bc$, where $a,b,c,d$ are positive integers.
We will prove the more general result that $a^n + b^n + c^n + d^n$ is composite whenever $ad = bc$ through a simple, but powerful argument through proportion.\Let, \ We choose $x$ such that $x \geq 1$. (This is always possible, because if it weren't then $a$ and $d$ would both be less than $b$ and $c$, which contradicts the equality.)
\begin{align} \frac{a}{b} = &\frac{c}{d} = x \\ \implies a = bx \text{ and also} \implies c = dx \notag\\ \text{Now,} S &= a^n + b^n + c^n + d^n \notag\\ &= (bx)^n + b^n + (dx)^n + d^n \notag\\ &= b^n(x^n + 1) + d^n(x^n + 1)\notag \\ &= (x^n + 1)(b^n + d^n) \end{align} $S$ is a product of two positive integers greater than $1$. Therefore, it is not prime. Hence, proved.
My doubt is if $S$ is necessarily a product of two integers greater than one. What if $b$ and $d$ are less than one ?