Let us say that polynomial $P(x)$ is "friendly" if it is an integer constant, or if it has all distinct integer roots and its derivative is friendly.

While fooling around with "friendly" polynomials I found it would be useful to characterize all integer solutions to

$$ m_1^2 + m_2^2 + m_3^2 - m_1 m_2 - m_2 m_3 - m_3 m_1 = n^2 $$

There are people out there who are better in number theory problems of this sort than I am, so I am asking for help.

What I have tried: I have tried the equation in various small-integer moduli, to see if one of them would provide a constraint on the forms of the $m_i$ and $n$. But I got no real progress there.



The equation is equivalent to $$(m_1 -m_3)^2 + (m_2 - m_3)^2 - (m_1 - m_3)(m_2 - m_3) = n^2$$

Substitute and get the equivalent equation $$a^2 + b^2 - a b = n^2 $$

It it simpler ( and more or less equivalent) to search for rational solutions of the equation $$x^2 + y^2 -x y = 1$$

We have a particular solution $(x,y) = (-1,-1)$. To get a general solution in rationals, use the substitution $y+1 = t(x+1)$, plug into the equation and solve for $x$, $y$ in terms of $t$. We get $$x = \frac{2 t - t^2}{t^2 -t + 1} \\ y = \frac{2 t-1}{t^2 -t+ 1} $$

This gives all the rational solutions of the equation. From here we conclude that the primitive solutions of the equation $a^2 - a b + b^2 = n^2$ must be of form $$a = 2 p q - p^2\\ b= 2 p q - q^2 \\ n= p^2 - p q + q^2$$ where $p$, $q$ relatively prime integers.

  • $\begingroup$ In this case, yes. One of the easiest examples where stereographic projection does not show all primitive integer solutions is $x^2 + 6 y^2 = z^2.$ For this example, you need to add just one extra parametrization of the same general type as your $(a,b,n)$ above. $\endgroup$ – Will Jagy Sep 14 '17 at 19:47
  • $\begingroup$ First $x = 6 p^2 - q^2,$ $y = 2pq,$ $z = 6 p^2 + q^2$ comes from projection. But then we need $x = 3 p^2 - 2 q^2,$ $y = 2 p q,$ $z = 3 p^2 + 2 q^2$ $\endgroup$ – Will Jagy Sep 14 '17 at 19:51
  • $\begingroup$ @Will Jagy: Thank you for the input. Are you saying the above parametrization misses some solution? I haven't check carefully the details, so it's possible. $\endgroup$ – Orest Bucicovschi Sep 14 '17 at 19:53
  • $\begingroup$ @Will Jagy: That is a good point. I more or less brushed over the "passing between rational and integer solution" step, one day I have to pay some close attention to it. $\endgroup$ – Orest Bucicovschi Sep 14 '17 at 19:54
  • 1
    $\begingroup$ @Will Jagy: that is very interesting, will try it. $\endgroup$ – Orest Bucicovschi Sep 14 '17 at 20:03


Solution we write.






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.