<s>[INST] Prove that if $n(a^2+b^2+c^2)=abc$ then $2\mid n$ 
Is it true that if $n\in\mathbb N$ and the diophantine equation $$n(a^2+b^2+c^2)=abc,\\(a,b)=(b,c)=(c,a)=1\tag1$$ 
  has positive integer solutions $a,b,c$, then $2\mid n$?

I can prove that $3\mid n:$
1) If $3\not\mid abc$ then $3\mid a^2+b^2+c^2,$ a contradiction.
2) If $3\mid abc,$ since $(a,b)=(b,c)=(c,a)=1$, we can assume that $3\mid a$ and $3\not \mid bc,$ then $3\not\mid a^2+b^2+c^2,$ hence $3\mid n.$
I can prove that equation $(1)$ has infinitely many solutions when $n=6,$  in fact, let $c=17,$ then it become a Pell's equation: $(12a-17b)^2-145b^2=-41616.$
I find some solutions to equation $(1)$:
$\{a,b,c,n\}=\{39,20,17,6\}\{52,29,15,6\}\{68,61,45,18\}\{87,80,61,24\}$
However, I cannot prove that $2\mid n.$ Thanks in advance!
 [/INST] 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, $2\mid n$.
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