Find all ordered pairs $(m,n)$ where $m$ and $n$ belongs to set of natural numbers,such that: $m$ divides $n^2+1$ and $n$ divides $m^2+1$

Question

I am struggling with a problem of elementary number theory. The problem is as follows:

Find all ordered pairs $(m,n)$ where $m$ and $n$ belongs to set of natural numbers,such that: $m$ divides $n^2+1$ and $n$ divides $m^2+1$.

I have not been able to do much but I am sharing what I have figured out:

i) $\gcd(m,n)=1$

ii) there exist $t$ such that $mnt=m^2+n^2+1$

iii) $t$ is a multiple of $3$.

Please provide me with some hints to solve the problem. Or a complete solution is most welcomed.

You have been around for a year and a half. Haven't you yet noticed that you are supposed to use MathJax around here? — José Carlos Santos, Jul 19 '20 at 10:50
In this link https://math.stackexchange.com/questions/115272/let-x-and-y-be-positive-integers-such-that-xy-mid-x2y21 , it is shown in fact that $t=3$. — GreginGre, Jul 19 '20 at 11:26

score 2 · Answer 1 · answered Jul 19 '20 at 11:28

Note that if $(m,n)$ is a solution with $m<n$, then $(\frac{m^2+1}n,m)$ is also a solution, with $\frac{m^2+1}n \leq m$ and both numbers decreasing.

This means that you will eventually descend to the solution $(1,1)$. Since the procedure is reversible, every solution is obtained by repeatedly applying the inverse: $(m,n)\mapsto(n, \frac{n^2+1}m)$ to the previous solution, starting from $(1,1)$.

Thus the first several are: $(1,1), (1,2), (2,5), (5, 13),\dots$.

I leave to you to figure out a general formula relating to Fibonacci numbers.

Find all ordered pairs $(m,n)$ where $m$ and $n$ belongs to set of natural numbers,such that: $m$ divides $n^2+1$ and $n$ divides $m^2+1$

1 Answers1