Find $m, n$ such that $\frac{n^2 + 1}{m^2 + 1 }$ is an integer multiple of a perfect square

Question

I'm trying to find $n,m \in \mathbb{N}$ such that $\sqrt{ \frac{n^2+1}{2(m^2+1)}}$ is rational.

I see that if $a,b$ are relatively prime $\sqrt{ \frac{a}{b}}$ is rational if and only if $a,b$ are perfect squares. $n^2+1$ can be a perfect square only if n = 0 and $2(m^2+1)$ is a square only when m = 1. For any other solution, we must have that WLOG $a = b\cdot r^2$ for some integer $r$. In other words $\frac{n^2+1}{m^2+1} = 2 \cdot r^2$

How could I go about solving that - or, what seems more likely, showing that there are no solutions? Is there a more general way to show $\frac{n^2+1}{m^2+1} = d \cdot r^2$ can have solutions only for specific $d$.

Thank you

There are various solutions for different $d$. For example $n=7, m = 3, d = 5$. — Calvin Lin, Jul 14 '13 at 21:22
@CalvinLin thanks, that's nice for $d=5$, but right now I'm after $d=2$ and the actual theory/approach to this kind of a problem — scibuff, Jul 14 '13 at 21:34
@scibuff: You can get lots of solutions by looking at the Pell Equation $x^2-2y^2=1$ and relatives. — André Nicolas, Jul 14 '13 at 21:38
$\frac{ (18^2 + 1 ) } { 2(5^2 + 1) } = 2.5^2$, so there are solutions for $d=2$ (at least for the rational case which you mentioned in your question). — Calvin Lin, Jul 14 '13 at 21:52
Maybe insert the multiplier 2 in the denominator of the title to the question, as it appears in the body of the question... — coffeemath, Jul 14 '13 at 22:53

Calvin Lin · Accepted Answer · 2013-07-14T22:26:13.707

2

As leshik points out, there are infinitely many solutions to

$$n^2 - 2m^2 = 1$$

Each of these gives $\frac{n^2 + 1 } { 2(m^2 + 1) } = 1$

I'm not certain if requiring an integer perfect square is possible, but your question states that rational numbers are fine.

Consider the Pell's equation $x^2 - 13 y^2 = -1$. It has a solution $(18, 5)$, hence has infinitely many solution.

Observe that for each solution,

$$\frac{x^2+1} { 2 (5^2 + 1) } = \left( \frac{y}{2} \right) ^2 $$

Hence, we have infinitely many pairs of integers of the form $(x, 5)$ which work.

Since the smallest solution of $x^2- 13 y^2 = 1$ is $(649, 180)$, the next solution is quite large, and is $(x,y) = (23382, 6485)$.

This arose from realizing that we have a solution $(18,5)$. To generalize this approach for any $d$ would require finding a specific solution first, before knowing what $D$ (in Pell's Equation) to use.

edited Jul 14 '13 at 22:26

answered Jul 14 '13 at 21:58

Calvin Lin

68,864

(1) Sweet, that makes sense. (2) How did you go about finding (18,5)? – scibuff Jul 14 '13 at 22:09
@scibuff Trial and error, by listing out values of $n^2+1$, which was pretty quick in this case. A general $d$ will be difficult through. – Calvin Lin Jul 14 '13 at 22:13
if you allow to use Pell's equation, then $n^2+1=2(m^2+1)$ has infinitely many solutions as well. – leshik Jul 14 '13 at 22:19
@leshik Damn, that's much better. $(3,2)$ works immediately (and I didn't check small solutions assuming that OP did). You should write up your solution. – Calvin Lin Jul 14 '13 at 22:24

Find $m, n$ such that $\frac{n^2 + 1}{m^2 + 1 }$ is an integer multiple of a perfect square

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