# Find all integers that make this expression rational

I came up with this difficult problem a while ago while solving another relatively easy problem.

Find all integers m and n, such that $m^2 + n^2$ is a square, and such that $\sqrt{\frac{2m^2+2}{n^2+1}}$ is rational.

I've already tried the pythagorean substitution numerous times to no avail and to make things worse it isn't even homogenous. Come to think of it I must have made a mistake in reducing my problem to this.

• First thing I notice: This is symmetric, since $\alpha=\sqrt{\frac{2m^2+2}{n^2+1}}$ is rational if and only if $\frac{2}{\alpha}=\sqrt{\frac{2n^2+2}{m^2+1}}$ is rational. – Thomas Andrews Dec 19 '14 at 1:01
• How about giving the original problem, then, rather than this one? – Thomas Andrews Dec 19 '14 at 1:01
• Because it's really easy and its from a competition whose solutions are made public? – Faraz Masroor Dec 19 '14 at 2:09
• If it's really easy, why is this problem so hard? Anyway, that's some chutzpah. – Thomas Andrews Dec 19 '14 at 2:29
• @FarazMasroor: Will you be interested if $m,n$ are rationals? Because just for info, the system, $$m^2+n^2 = x^2,\quad \frac{2(m^2+1)}{n^2+1}=y^2$$ is reducible to an elliptic curve, so there are infinitely many rational points $m,n$. – Tito Piezas III Dec 19 '14 at 23:37

Too long for a comment. As requested by OP as a sidenote, if we relax the requirements on $m,n$ and allow rationals, then the system,

$$m^2+n^2 = x^2,\quad \frac{2(m^2+1)}{n^2+1}=y^2\tag1$$

does have an infinite number of rational solutions. Other than the obvious case $n=1$, we have two quadratic polynomials in $m$ to be made squares,

$$f(m) = x^2,\quad\quad g(m) = y^2$$

hence an intersection of two quadric surfaces. A simple rational solution to $(1)$ is then,

$$m = \frac{(a-c)c}{(a-2c)b},\quad n=\frac{a}{2b}$$

where,

$$a^2+2b^2 = c^2\tag2$$

and $(2)$ is easily solved as $a,b,c = u^2-2v^2,\, 2uv,\, u^2+2v^2$ yielding,

$$x^2=\frac{(u^4+4u^2v^2+20v^4)^2}{(4uv)^2(u^2+6v^2)^2},\quad y^2 = \frac{2(u^2+4v^2)\,(4v)^2}{(u^2+6v^2)^2}$$

Thus, it remains to find $u^2+4v^2 = 2z^2$ which can also be solved parametrically. For a particular example, let $u,v,z = 2,\,1,\,4,$ so $m,n = \frac{3}{5},\frac{1}{4},$ and $x,y = \frac{13}{20},\frac{8}{5}$.

P.S. There are no positive integer solutions to $(1)$ with $m,n<500$.

This not an answer,but this may explain why there is no any answer and what are the issues this sort of problems arise, I hope this will be helpful for tracking the problem.

Given two integers $m$ and $n$ such that the two conditions holds,one can show-using the second condition- that there exists $p,d,q$ such that (up to permutation of $m$ and $n$):

\begin{align} m^2-dq^2&=&-1\\ n^2-2dp^2&=&-1 \end{align}

These two equation will have solutions in integers if $d$ satisfies some conditions ( $d$ odd non-square, $d$ is not divisible by any prime of the form $4k+3$, and others (which are more complicated see here ), Using this equations and knowing some initial solutions we can construct all integers $(m,n)$ verifying the two equations, we pick an element $d$ which verifies the two equations with the fundamental solution $(m_1,q_1)$and $(n_1,p_1)$ the solutions will be of the form $n_k$ and $m_{k'}$ where $k,k'$ are odd and: \begin{align} m_{k}+\sqrt{d}q_{k}&=&(m_1+\sqrt{d}q_1)^k\\ n_{k}+\sqrt{2d}p_{k}&=&(n_1+\sqrt{2d}p_1)^k \end{align}

asking whether $m_k^2+n_{k'}^2$ is a square is generally,a very diﬃcult and we do not have any general theory which could be used even when we know $d$,we don't even know when for example $n_k$ is a square.

we can also define the sequence $m_k,n_k$ by recurrence : \begin{align} m_{k+2}&=2m_1 m_{k+1}+m_k, &m_0=1\\ n_{k+2}&=2n_1 n_{k+1}+n_k, &n_0=1\ \end{align} This two sequence are almost Fibonacci polynomials, which generalize Fibonacci numbers, they are usually noted $F_k(2m_1),F_k(2n_1)$, the question of whether $F_k(a)^2+F_{k'}(a)^2$ is an open problem and conjectured to be finite (see here), and our question is more general which is $F_k^2(2n_1)+F_{k'}^2(2m_1)$ is a square?.

I think that this is an analyse of your question, but sometimes there exists shortcuts we don't know.