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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!

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

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