Prove that if $n(a^2+b^2+c^2)=abc$ then $2\mid n$

Solution 1:

It is indeed true. We need the following well known fact:

If $p\equiv3\pmod4$ is prime and $p\mid x^2+y^2$, then $p\mid x$ and $p\mid y$.

We will prove that

Theorem. If $a,b,c$ is a solution to $(1)$, then exactly one of $a,b,c$ is divisible by $4$.

Proof.

First suppose $a,b,c$ are all odd. Then $a^2+b^2+c^2\equiv3\pmod 4$, so has a prime divisor $p\equiv3\pmod 4$. Without loss of generality, suppose $p\mid a$.

Then $p\mid b^2+c^2$, so $p\mid b$ and $p\mid c$ which contradicts the condition $(a,b)=(b,c)=(c,a)=1$.

Therefore one of $a,b,c$ is even. Say $2\mid a$ and suppose $4\nmid a$. Then $a^2+b^2+c^2\equiv6\pmod8$, which means it has a prime divisor $p\equiv3\pmod 4$. Again $p\mid b$ and $p\mid c$, contradiction.

So we should have $4\mid a$. $\square$

In this case, $a^2+b^2+c^2\equiv2\pmod 4$, which means $4\nmid a^2+b^2+c^2$. Therefore, $2\mid n$.