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Let $A$ be a set of all positive integers so that if $ n\in A $, then $n-1$ has at least one prime divisor $p\equiv 2( mod 3)$ such that $v_p(n-1)$ is odd. And let $B$ be a set of all positive integers so that if $n\in B$ then $n+2$ has at least one prime divisor $p\equiv 2( mod 3)$ such that $v_p(n+2)$ is odd.

Let $k$ be a positive integer that does not belong to neither $A $ nor $B$.

Prove that the equation $ x^2+y^2+z^2=k(xy+yz+zx)$ has a solution in positive integers.

amWhy
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user140378
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  • Strange collection of conditions. Have you tested this for all $n$ up to whatever? Where does this question come from? – Gerry Myerson May 20 '15 at 07:32
  • "Some examples say yes...." How many examples? A thousand? A million? Is there a proof written down somewhere that there are no nontrivial solutions for $n$ in those sets? Is there a proof that there's a solution, if you don't restrict to positive integers? The more you tell people, the easier it will be for someone to help you. – Gerry Myerson May 20 '15 at 09:10
  • No, I didnt check thousand examples. – user140378 May 20 '15 at 09:12
  • What then? Three examples? – Gerry Myerson May 20 '15 at 09:13
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    http://www.artofproblemsolving.com/community/c6h1064861p4622154 – user140378 May 20 '15 at 09:17
  • n=5 : (3,5,41) , n=14: (2,3,71), – user140378 May 20 '15 at 14:09
  • So to somplify the condition: $n$ is an integer such that for both $n-2$ and $n+1$ all primes $\equiv2 \pmod 3$ occur in even power. I suppose that menas thet $n-2$ and $n+1$ are norms in $\mathbb Z[\rho]$. – Hagen von Eitzen May 21 '15 at 20:07
  • The significance of $n-1$ and $n+2$ (sign error in my last comment) may arise from $(x+y+z)^2=(n+2)(xy+xz+yz)$ and $(x-y)^2+(y-z)^2+(z-x)^2=2(n-1)(xy+yz+xz)$. – Hagen von Eitzen May 22 '15 at 06:30

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