<s>[INST] Prove that $a^2  + b^2 + c^2 $ is not a prime number I am having difficulty solving this problem:
Let $a, b, c \in\mathbb{Z}$, $abc \neq 0$ and $a\neq c$ be such that
$$\frac{a}{c} = \frac{a^2+b^2}{c^2+b^2}.$$
Prove that $a^2 + b^2 + c^2$ is not a prime number.
Thanks in advance!
 [/INST] W.l.o.g we may assume that $a,c>0$. The equation
$$
\frac{a}{c}=\frac{a^2+b^2}{c^2+b^2}
$$
together with the assumption $a\neq c$ quickly gives us $ac=b^2$ as a corollary. Therefore we have
$$
a^2+b^2+c^2=a^2+ac+c^2
$$
and the extra condition that $ac=b^2$ must be a perfect square.
There are two main cases. If $gcd(a,c)>1$, then that common divisor is  also a divisor of $a^2+ac+c^2$, so this latter number won't be a prime. If the numbers $a$ and $c$ are coprime, then the equation $ac=b^2$ and unique factorization force both $a$ and $c$ to be squares. So we can assume that $a=p^2, c=q^2$ for some integers $p,q$. But then we see that
$$
a^2+b^2+c^2=p^4+p^2q^2+q^4=(p^2+q^2)^2-p^2q^2=(p^2+pq+q^2)(p^2-pq+q^2).
$$
Here $p\neq q$, so both these factors are $>1$, and the claim follows in this case, too.
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