Are there infinitely many pairs $r, s \in \mathbb{Q}$ such that $r + s \neq 0$ and $\frac{1}{r} + \frac{1}{s}$ are both integers?

Question

If we restrict $r, s$ to only $\mathbb{Z}$, it's easy to prove that the only possibilities are $r = s \in \{1,2 \}$, but I haven't been able to prove this more general case.

Also, is it significantly harder to answer this questions if, say, $r, s \in \mathbb{R}$?

(as pointed out in the comments, there's a trivial solution to the problem... the question becomes much more interesting without that, so I've edited the question)

Answer 1

Let $r+s=m$, $\frac1r+\frac1s=n$. So we want $$ nr^2-mnr+m=0 $$ So we want $\Delta=(mn)^2-4mn=(mn-2)^2-4$ to be a perfect square (in $\mathbb{Q}$ or in $\mathbb{R}$). For $\mathbb{Q}$, integer perfect squares are of course square of integers, so we must have $mn-2=\pm 2$ which yields the trivial solutions $mn=0$, or $mn=4$, which are $r=s\in\{\pm 1,\pm 2\}$.

For $\mathbb{R}$, we have much more freedom. Any $mn\neq 1,2,3$ (and $mn\neq 0$ to rule out trivial solutions) would work, as $\Delta\geq 0$, we can take its square-root $\sqrt{\Delta}\in\mathbb{R}$ and obtain: $$ r=\frac{mn\pm\sqrt{\Delta}}{2n},s=\frac{mn\mp\sqrt{\Delta}}{2n} $$

Answer 2

$$ (r+s) \left( \frac{1}{r} + \frac{1}{s} \right) = 2 + \frac{r}{s} + \frac{s}{r} \in \mathbb Z $$ Let $$ x = \frac{r}{s} \in \mathbb Q $$ $$ x + \frac{1}{x} = n \in \mathbb Z $$ $$ x^2 - n x + 1 = 0 $$ $$ x = \frac{n \pm \sqrt{n^2 - 4}}{2} \in \mathbb Q $$ $$ n = \pm 2 $$

Answer 3

Here's a completely elementary solution (without using the formula for quadratic roots).

If $r+s$ and $\frac 1 r+\frac 1 s$ are both integers, then $\frac r s+\frac sr+2=(r+s)(\frac 1 r+\frac 1 s)$ is also an integer, and hence $\frac r s+\frac s r$ is an integer.

Write $\frac{r}{s}$ as $\frac{p}{q}$, where $p,q$ are coprime integers, so that $\frac r s+\frac s r=\frac pq+\frac qp=\frac{p^2+q^2}{pq}$. If this is an integer, it means that $p$ has to divide $q^2$ and $q$ has to divide $p^2$. But since $p$ and $q$ are coprime, it follows that they have no prime factors, and so each is equal to $\pm 1$, and so $\frac{r}{s}=\frac{p}{q}=\pm 1$, and hence $r=s$ or $r=-s$.

You specifically excluded $r=-s$, so the only remaining solutions have $r=s$, i.e. $r+s=2r$ is an integer and $\frac 1r+\frac 1s=\frac 2r$ is an integer. The only $r$ satisfying this are $-2,-1,-\frac{1}{2},\frac{1}{2},1,2$.

(This solution does not work at all if you allow real $r,s$, as in that case there are infinitely many $x$ such that $x+x^{-1}$ is an integer. But as mentioned in the comments, in that case, we can explicitly define infinitely many solutions.)

Answer 4

Let $r+s = m;m\in \mathbb Z$ and $\frac 1r + \frac 1s = \frac 1r + \frac 1{m-r} = \frac {(m-r) + r}{r( m-r)} = \frac m{r(m-r)} = k; k \in \mathbb Z$.

$m = kr(m-r)$ and $kr^2 - kmr - m = 0$ so $r = \frac {km \pm \sqrt{k^2m^2 +4mk}}{2k} \in \mathbb Q$

So for any two integers were $k^2m^2+ 4mk$ is a perfect square with have a solution. So $mk(mk + 4)= w^2$. Now the $\gcd(mk, mk+4) = \gcd(4, mk)$

Case 1: $\gcd(mk,mk+4) = 1$ then $mk$ and $mk+4$ are perfect squares and the only way that can be is if $mk = 0$ and $mk + 4 = 4$. So any $mk = 0$ so $r=-s$ is possible.

Case 2: $\gcd(mk,mk+4) =2$ and $\frac {mk}2(\frac {mk}2 + 2) = (\frac w2)^2$ and $\frac {mk}2 $ and $\frac {mk}2 + 2$ are perfect squares. The only way that is possible is if $\frac {mk}2 = -1$ and $\frac {mk}2 = 1$. That can only happen if $\{m,k\} = \pm 1, \mp 2$

But that's impossible as $r =\frac {km \pm \sqrt {k^2m^2 + 4km}}{2k} = \frac {-2 \pm \sqrt{4-8}}{2}$.

Case 3: $\gcd(mk,mk+4) =4$ and $\frac {mk}4(\frac {mk}4 + 1) = (\frac w4)^2$. where $\frac {mk}4$ and $\frac {mk}4 + 1$ are perfect squares. That's only possible of $mk = 0$ and $\frac {mk}4 + 1 =1$.

And that would mean $r = -s$.

So $r = -s$ are only solutions.

At least if I didn't completely screw up.... Which I might have.