Let $\Bbb S$ be the set of rational numbers $r$ with the property that $\sqrt{r}$ is rational as well, in other words a number $\frac{a^2}{b^2}$ with integers $a,b$ and $b\ne 0$
Example is $\frac{81}{121}=(\frac{9}{11})^2$
My question is this: Given some $x\in\Bbb Z$, are there infinitely many $\mu \in \Bbb S$ for which $(x+\mu)\in \Bbb S$? And if so, is there a general formula I can use for computing these?
I'm using this idea to approximate $\sqrt{x}$ rationally and want to see if there is an easy way to find the solutions
In my research, the best I found was via this method:
$a=(x-1)^{2^{n-1}}$
$b$ is the $2^n$th result of $a_n=2a_{n-1}+(x-1)a_{n-2}$ - searchable on OEIS using the first five terms $\big[0,1,2,(x+3),(4x+4)\big]$
E.g for $x=3$, we use http://oeis.org/A002605, of which the $2^3$rd term is $896$, and $a=2^4=16$
$$\sqrt{3+\bigg(\frac{16}{896}\bigg)^2}=\frac{97}{56}\approx\sqrt3+0.00009$$ Using the $2^4$th term for $x=3$ we would get $a=256, b=2781184$: $$\sqrt{3+\bigg(\frac{256}{2781184}\bigg)^2}=\sqrt{3+\bigg(\frac{1}{10864}\bigg)^2} = \frac{18817}{10864}\approx\sqrt{3}+(2\times10^{-9})$$
What I mainly want to know is if there are other similar methods I can use to find other approximations.
So, you've been given infinitely many solutions in the comments: if $x\in \mathbb{Z}$, then for any $p/q\in\mathbb{Q}$, $\mu(x, p/q)=\left(\frac{xq^2-p^2}{2pq}\right)^2\in\mathbb{S}$, and moreover $$ x + \mu(x,p/q)=x+\left(\frac{xq^2-p^2}{2pq}\right)^2=\frac{4p^2q^2x+(x^2 q^4 - 2p^2 q^2 x + p^4)}{4p^2q^2}=\left(\frac{xq^2+p^2}{2pq}\right)^2=\mu(-x,p/q), $$ which is in $\mathbb{S}$ as well. If you'd additionally like the chosen $\mu$ to be small (which isn't specified in the question, but is implied by the bounty), then take $p/q$ to be approximately $x$. For instance, take $p=qx - 1$; then $$\mu(x,p/q)=\left(\frac{xq^2-(qx - 1)^2}{2(qx - 1)q}\right)^2=\left(\frac{2qx-1}{2(qx-1)q}\right)^2\rightarrow 0$$ as $q\rightarrow\infty$.