Find the maximum value of $m^2+n^2$ if $(m^2-mn-n^2)^2=1$

Question

Given that the integers $m$ and $n$ in the set $A=\left\{1,2,3,....,2024\right\}$ satisfy $(m^2-mn-n^2)^2=1$. Find the maximum possible value of $m^2+n^2$.

My effort: We have $m^2-mn-n^2=\pm 1$

Case $1.$ If $m^2-mn-n^2=1 \Rightarrow m^2-mn-(n^2+1)=0$

Now the Discriminant is $$D=n^2+4(n^2+1)=k^2, k \in \mathbb{Z}$$

$$ \implies 5n^2+4=k^2$$ I am not able to proceed now.Same problem with Case $2.$

Answer 1

If $(m^2-mn-n^2)^2=1$ then $$(2m-n)^2-5n^2=4m^2-4mn-4n^2=\pm4,$$ which is a (pair of) Pell equations. For $(m,n)=(1,0)$ and $(m,n)=(0,1)$ we have $$(2\cdot1-0)^2-5\cdot0^2=4\qquad\text{ and }\qquad (2\cdot1-1)^2-5\cdot(-1)^2=-4,$$ and the fundamental solution is $2+\sqrt{5}$. A few quick computations show that \begin{eqnarray} 2\cdot(2+\sqrt{5})^5\ \ &=&\ \ 1364+\ \ 610\sqrt{5},\\ 2\cdot(2+\sqrt{5})^6\ \ &=&\ \ 5778+2584\sqrt{5},\\ (1+\sqrt{5})\cdot(2+\sqrt{5})^5\ \ &=&\ \ 2207+\ \ 987\sqrt{5},\\ (1+\sqrt{5})\cdot(2+\sqrt{5})^6\ \ &=&\ \ 9349+4181\sqrt{5},\\ (1+\sqrt{5})\cdot(2+\sqrt{5})^{-5}&=&\ \ \ \ 843-\ \ 377\sqrt{5},\\ (1+\sqrt{5})\cdot(2+\sqrt{5})^{-6}&=&-3571-1597\sqrt{5},\\ \end{eqnarray} and so the maximum is at $(m,n)=(1597,987)$.

Answer 2

Continuing from your relaxed equation,

$m^2 - mn - n^2 = \pm 1$

$m^2 - mn - n^2 \pm 1 = 0 $

m = $ \frac{n \pm {\sqrt {5n^2 \pm 4}}}{2} $

As a property, x is a fibonacci number if atleast one of $5x^2 + 4$ and $5x^2 - 4$ is a perfect square.

So, for a valid m to exist, $5n^2 \pm 4$ must be a perfect square, and thus a/c to the lemma, n must be a fibonacci number.

What are possible values of n if it's a fibonacci number ?

$1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597$ (since all of these lie in the range $1,2,...2024$).

Putting n = $1597$ into the solution, results m = $2584$ which isn't possible since m lies out of the permissible range, i.e. $(1,2,...2024)$.

Putting n = $987$ into the solution, results m = $1597$ and these will be the maximum values of n & m that you can achieve since you began with the best values of n.

Hence, the answer to your question $m^2 + n^2$ = $1597^2 + 987^2$.