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