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!

  • $\begingroup$ observation: this is the same as proving exactly one of $a,b,c$ is even. $\endgroup$ – Guy Apr 2 '14 at 14:25
  • $\begingroup$ @Sabyasachi I wonder how to rule out that $a,b,c,n$ are all odd. $\endgroup$ – Next Apr 2 '14 at 14:27
  • $\begingroup$ Have you tried taking a,b,c as of form 2k+1 etc and then look at the behaviour of both sides? $\endgroup$ – O_huck Apr 2 '14 at 14:31
  • 1
    $\begingroup$ @Jérémy Blanc Well, he only asked me if equation $(1)$ has integer solutions, I found some solutions, so I have answered his questions. But I have a hobby, when I saw a problem, I would think what can be discussed from this problem. When I solve $(1)$, I found that for all the solutions I can find, $n$ are even, but I cannot prove that, so I ask this question. $\endgroup$ – Next Apr 2 '14 at 15:28
  • 1
    $\begingroup$ If we remove $(a,b)=(b,c)=(c,a)=1$ then $n=9,a=21,b=35,c=42$ is a solution to $(1)$, I think we cannot prove it only by $\mod something$. $\endgroup$ – Next Apr 2 '14 at 15:44

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 $4$.


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, $2\mid n$.

  • 2
    $\begingroup$ Well done! Thanks:) $\endgroup$ – Next Apr 9 '14 at 11:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.