# Checking proof of simple number theory problem

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 ?

• who said $x$ needs to be an integer ? Jul 4, 2016 at 7:16
• @avz2611 Youre right. Perhaps, I could expand $x$ in terms of the original ratio. Jul 4, 2016 at 7:29

Let $g$ be the greatest common divisor of $a$ and $b$. Then we can write $$a=gs\,,\ b=gt\,,\ g,s,t\in{\Bbb Z}^+\,,\ \gcd(s,t)=1\,.$$ Then $ad=bc$ becomes $$sd=tc\ ;$$ since $s,t$ are coprime we have $s\mid c$, say $c=su$, and substituting back gives $$c=su\,,\ d=tu\,,\ u\in{\Bbb Z}^+\,.$$ Hence $$a^n+b^n+c^n+d^n=(s^n+t^n)(g^n+u^n)$$ which is composite.