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.
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.