Find integers $f,g,h$ such that $3(f^2+g^2+h^2)=14(fg+gh+hf)$.

You can do it using a computer or by hand.

I tried this problem for ages, got nowhere. Unfortunately I don't know how to program, but I thought it would help a lot here. (Just set up a program to check loads of values until you get one that works?)

I would appreciate any type of solution to this. I only need one (f,g,h) that works.


  • $\begingroup$ If you allow all integer values for $f,g,h$, I'd recommend going for the quick $(0,0,0)$-solution, but I imagine that's not exactly what you were looking for. Before I spend a lot of time on finding a solution, do you know that a non-trivial solution exists? Because depending on whether you are certain, that would change the approach to the problem a lot. $\endgroup$ – Some Math Student Feb 4 '15 at 20:36
  • $\begingroup$ Exhaustive computer search finds that there is no solution with $0\le f,g,h\le 2000$ other than the obvious $f=g=h=0$. $\endgroup$ – MJD Feb 4 '15 at 20:48

We have $3(f+g+h)^2 = 3(f^2 + g^2 + h^2) + 6(fg + gh + hf) = 20(fg + gh + hf)$, so that both $f+g+h$ and $fg + gh + hf$ are divisible by $5$.

Plugging in the roots of $(X-f)(X-g)(X-h) = X^3 - (f+g+h)X^2 + (fg+gh+hf)X - fgh$, we find that $f^3 \equiv g^3 \equiv h^3 \equiv fgh\pmod{5}$, which (by uniqueness of cube roots modulo $5$) is only possible if $f\equiv g \equiv h\pmod{5}$, and it follows that $f$, $g$, and $h$ are all divisible by $5$.

But $(f/5,g/5,h/5)$ would then give us another solution of smaller magnitude. By descent, the only possible solution is $(0,0,0)$.

  • $\begingroup$ @AnalysisIncarnate Is there a typo? $n=2$ doesn't give the same problem ($19$ instead of $14$). $\endgroup$ – Slade Feb 4 '15 at 21:16
  • $\begingroup$ @AnalysisIncarnate Ah, I see, this is the case $n=3$. $\endgroup$ – Slade Feb 4 '15 at 21:17
  • $\begingroup$ I'm not following part of your argument. What do you mean by "plugging in the roots"? $\endgroup$ – MJD Feb 4 '15 at 21:18
  • $\begingroup$ @MJD Setting $X=f,g,h$. $\endgroup$ – Slade Feb 4 '15 at 21:20
  • 3
    $\begingroup$ @AnalysisIncarnate Cube roots are unique mod $5$, so $f^3\equiv g^3 \equiv h^3 \implies f\equiv g \equiv h$. Since $f+g+h\equiv 0$, we must have $f\equiv g\equiv h \equiv 0$. $\endgroup$ – Slade Feb 4 '15 at 21:29

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