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I've been trying to solve this for hours and got nowhere, so I can only assume it's a really difficult problem.

Problem: Find polynomials $f,g,h$ with integer coefficients such that:

$(3n-3)(f^2+g^2+h^2)=(9n+1)(fg+gh+hf)$,

where $n$ is a positive integer.

Thanks

Ninja Boy
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    At least for some $n$ there are no nontrivial solutions. Plug in $0$ for $x$. Then for the constant terms $a, b, c$ of $f, g, h$ we must have $(ab+bc+ca)/(a^2+b^2+c^2)$ must be equal to $(3n-3)/(9n+1)$ (if $a^2+b^2+c^2 \neq 0$). This is impossible, for example, for $n=3$. So $a^2+b^2+c^2=0$. So $a, b, c=0$. Divide everything through by $x$ and repeat. So $f, g, h=0$. – Marcel Besixdouze Feb 04 '15 at 20:28
  • A solution is given by $n=1$, $f=g=2$, $h=-1$. – Gerry Myerson Feb 14 '15 at 09:33
  • Are you still here, Analysis? – Gerry Myerson Feb 15 '15 at 11:05
  • @GerryMyerson, as near as I can make out Analysis and Ark are the same person, along with Eulerian Adventure. It turned out that, given integers $B > A > 0$ and $\gcd(A,B) = 1,$ there is an integer solution to $$ A(x^2 + y^2 + z^2) = B (yz+zx+xy), $$ with $x,y,z$ not all zero, if and only if both $B+2A$ and $B-A$ are represented by the binary form $u^2 + 3 v^2.$ When this holds, there is a recipe for $x,y,z.$ See http://math.stackexchange.com/questions/1148380/when-does-ax2y2z2-bxy-yz-xz-have-nontrivial-integer-solutions/1151011#1151011 – Will Jagy Feb 16 '15 at 22:28

1 Answers1

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Here is one specialized solution.

Let $h=g$. The the original equation becomes:

$$(3n-3)(f^2+2g^2)=(9n+1)(2fg+g^2)\tag{1}$$

We solve (1) for $f$ and obtain (assuming $g>0$): $$f=A(n)g\tag{2}$$ $$A(n)=\frac{9n+1\pm\sqrt{10(9n^2+3n-2)}}{3(n-1)}\tag{3}$$

(a) For $n=2$, We have $A(2)=-1/3,13$. If $g=3 j$ where $j$ is arbitrary positive polynomial with integer coefficients, then $(f,g,h=g)$ is a solution to the original equation.

(b) For $n=81$, We have $A(81)=-1/6,25/4$. If $g=12 j$ where $j$ is arbitrary positive polynomial with integer coefficients, then $(f,g,h=g)$ is another solution to the original equation.

Other solutions can be found out in similar fashion.

mike
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