# About the equation $m^4+n^4=10m^2n^2+1$

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.

• do you know pells equation? – Jorge Fernández Hidalgo May 31 '17 at 21:56
• It's $x^2−dy^2 = 1$ isn't it? – Kc2 May 31 '17 at 21:59
• Use the discriminant of the polynomial Why not try to use that hint? – dxiv May 31 '17 at 22:01
• 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. – Will Jagy May 31 '17 at 22:11
• can you unaccept so I can delete? – Jorge Fernández Hidalgo May 31 '17 at 22:15

## 1 Answer

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.

• 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. – Ricardo Largaespada Jun 5 '17 at 15:01
• @RicardoLargaespada you are right my friend. I should not said that it suffices to show. Sorry – Chito Miranda Jun 5 '17 at 19:29