https://math.stackexchange.com/questions/631198/why-cant-x2y2z2-be-prime?noredirect=1 https://artofproblemsolving.com/community/c4h1236079p6282199

parmenides51 wrote: Let $a, b, c$ be non zero integers,$a\ne c$ such that $$\frac{a}{c}=\frac{a^2+b^2}{c^2+d^2}$$ Prove that $a^2 +b^2 +c^2$ cannot be a prime number. (!) Instead of $d$ is $b$. The official solution translated by ChatGPT: **Problem 2.** We have: $$\frac{a}{c} = \frac{a^2 + b^2}{b^2 + c^2} \iff (a-c)(b^2 - ac) = 0.$$ But $a \neq c$ and thus, $b^2 = ac$. In this case, we have: $$a^2 + b^2 + c^2 = a^2 + 2ac + c^2 - b^2 = (a+c-b)(a+c+b).$$ Since $a^2 + b^2 + c^2$ is a natural number and $a^2 + b^2 + c^2 > 3$, if $a^2 + b^2 + c^2$ were a prime natural number, then we would have the following cases: 1. $a+c-b = 1$ and $a+c+b = a^2 + b^2 + c^2$, 2. $a+c+b = 1$ and $a+c-b = a^2 + b^2 + c^2$, 3. $a+c-b = -1$ and $a+c+b = -a^2 - b^2 - c^2$, 4. $a+c+b = -1$ and $a+c-b = -a^2 - b^2 - c^2$. In cases (1) and (2) we are led to: $$-2(a+c) + a^2 + b^2 + c^2 + 1 = 0 \iff (a-1)^2 + (c-1)^2 + b^2 = 1$$ and considering that $b \in \mathbb{Z}$, we obtain $b^2 = 1$ and $a = c = 1$. In cases (3) and (4) we have: $$2(a+c) + a^2 + b^2 + c^2 + 1 = 0 \iff (a+1)^2 + (c+1)^2 + b^2 = 1$$ and considering that $b \in \mathbb{Z}$, we obtain $b^2 = 1$ and $a = c = -1$. Therefore, in all four cases we contradict the hypothesis $(a \neq c)$, hence $$a^2 + b^2 + c^2$$ is a composite natural number.

parmenides51 wrote: Let $a, b, c$ be non zero integers,$a\ne c$ such that $$\frac{a}{c}=\frac{a^2+b^2}{c^2+\color{red}{d^2}}$$ Prove that $a^2 +b^2 +c^2$ cannot be a prime number. Isn't it $b^2?$

@above, indeed wording had a typo