Non-prime numbers

We have $\begin{aligned} (a^2+b^2)c & = \; a(b^2 + c^2) \\ ac(a-c) & = \; b^2(a-c) \end{aligned}$ and hence $b^2 = ac$ . Thus $a^2 + b^2 + c^2 \; = \; a^2 + 2b^2 + c^2 - b^2 \; = \; a^2 + 2ac + c^2 - b^2 \; = \; (a + c)^2 - b^2 \; = \; (a + c - b)(a + b + c)$ If $a+c-b=1$ then $a+c-1 = b = \sqrt{ac} \le \tfrac12(a+c)$ , so that $a+c \le 2$ . This is impossible with $a,c$ distinct positive integers. Thus we deduce that $a+c-b > 1$ , and hence $a^2 + b^2 + c^2$ is $\boxed{\mathrm{not\;\;prime}}$ .