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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.

Michael Hardy
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Rhys Hughes
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    $x+\left(\frac{x/y-y}{2}\right)^2=\left(\frac{x/y+y}{2}\right)^2$. –  May 02 '18 at 11:25
  • $\Bbb S$ isn't the non-negative rationals. In fact the vast majority of non-negative rationals aren't in $\Bbb S$ - it is any rational whose square root is also rational - like $\frac{1}{4}$, whereas for example $\frac{1}{2} \notin \Bbb S$ as $\sqrt{\frac{1}{2}}$ is irrational. – Rhys Hughes May 02 '18 at 12:04
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    @RhysHughes That comment is irrelevant. totoro has given you an element of your $\Bbb S$, namely $\mu=((xy-y)/2)^2$ for which $x+\mu\in \Bbb S$. Moreover he/she has given you a general formula for all solutions. – Angina Seng May 05 '18 at 11:29

2 Answers2

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You are asking, in effect if there are infinitely many rationals $a$ and $b$ with $$x+a^2=b^2.$$ This is the same as $$(b-a)(a+b)=x.$$ As long as $x\ne0$, you can make $b-a=c$, any nonzero rational, and $a+b=x/c$ etc.

Angina Seng
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  • Not quite. The equation would simply imply that $\frac{b^2x+a^2}{b^2}$ is a square number. We know that $b^2$ is a square number, but the numerator I believe is more awkward. – Rhys Hughes May 02 '18 at 11:51
  • @RhysHughes I disagree: I am not attempting to find the numerators and denominators separately; all my variables range over $\Bbb Q$. totoro's comment says the same thing. – Angina Seng May 04 '18 at 19:48
  • I see where you've gone with this, but the issue is that it doesn't exactly help me in terms of finding a and b in your case. You've answered one of the two questions here so I'm grateful for that, any thoughts on the second question? – Rhys Hughes May 05 '18 at 10:51
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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 $q\rightarrow\infty$.

mjqxxxx
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