Solve the diophantic equation: $$m^4 + n^4 = 10m^2n^2 + 1.$$

[Hint: Use the discriminant of the polynomial]

I did $m^4 - 10m^2n^2 + n^4 = 1$ I know that if $\gcd(x,y)\mid c $ this can be solved, but I don't know if I can do this step.

Thanks for your help.

  • $\begingroup$ do you know pells equation? $\endgroup$ – Jorge Fernández Hidalgo May 31 '17 at 21:56
  • 1
    $\begingroup$ It's $x^2−dy^2 = 1$ isn't it? $\endgroup$ – Kc2 May 31 '17 at 21:59
  • $\begingroup$ Use the discriminant of the polynomial Why not try to use that hint? $\endgroup$ – dxiv May 31 '17 at 22:01
  • $\begingroup$ You should know that there are infinitely many solutions to $u^2 - 10 uv + v^2 = 1,$ this may be familiar as "Vieta Jumping." There will be only finitely many solutions to your actual problem, correct treatment would involve factoring in $\mathbb Z [ \sqrt 6 ]$ I guess. $\endgroup$ – Will Jagy May 31 '17 at 22:11
  • $\begingroup$ can you unaccept so I can delete? $\endgroup$ – Jorge Fernández Hidalgo May 31 '17 at 22:15

Okay, I'll try. The equation can be rewritten as $(m^2-n^2)^2-2(2mn)^2=1$. So, letting $x=m^2-n^2$ and $y=2mn$, we get the equation $x^2-2y^2=1.$ Hence, to get the solutions of the original equation, it suffices to solve the diophantine equation $x^2-2y^2=1$. But this new equation is precisely the question of finding the units in the ring $\mathbb{Z}[\sqrt{2}]$ with norm $1$. Note that the fundamental unit of this ring is $1+\sqrt{2}$ which correspond to $x=1$ and $y=1$. This fundamental unit has norm $-1$. In fact, the units in the ring $\mathbb{Z}[\sqrt{2}]$ are of the form $\pm(1+\sqrt{2})^k$, where $k$ is an integer. So the units of norm $1$ are those units $\pm(1+\sqrt{2})^k$ where $k$ is even.

| cite | improve this answer | |
  • 2
    $\begingroup$ But is not true that an solution for $x^2-2y^2=1$ is also solution for the original problem, i.e. $x=3$, $y=2$ implies $m^2-n^2=3$ and $2mn=2$ which do not have solutions in integers. $\endgroup$ – Ricardo Largaespada Jun 5 '17 at 15:01
  • $\begingroup$ @RicardoLargaespada you are right my friend. I should not said that it suffices to show. Sorry $\endgroup$ – Chito Miranda Jun 5 '17 at 19:29

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.