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I would like to find the positive integers $k$ for which $\sqrt{1+\frac{k}{n}}$ is irrational for all $n\in\mathbb{N}$.

I was led to this question when I was making up an example for my class, and I suspect that this is only true for a few small values of $k$.

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Suppose $k\ge3$ is a multiple of an odd number. Then there exist $a,b\in\mathbb{N}$ such that $k=(2a+1)b.$ For $n=a^2b,$ $$\sqrt{1+\dfrac{(2a+1)b}{a^2b}}=1+\dfrac{1}{a}\in\mathbb{Q}$$ Now we have only consider about even $k$ values. Suppose $k$ is a multiple of $8.$ Then there exist $a\in\mathbb{N}$ such that $k=8a.$ For $n=a,$ $$\sqrt{1+\dfrac{8a}{a}}=3\in\mathbb{Q}$$

Then only we have remain the cases $$k=1,2,4$$ which can be treated individually.

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    $\begingroup$ Oh, this answer is much better than mine! $\endgroup$ – Barry Cipra Oct 30 '14 at 16:39
  • $\begingroup$ Thanks for your answer. Is it easy to verify that $k=1,2,4$ all satisfy this property? $\endgroup$ – user84413 Oct 30 '14 at 16:52
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    $\begingroup$ Nice work to @Barry and Nilan. It is pretty easy to show that $k=1,2,4$ all work. For example, for $k=1$, you need to show that $(n+1)/n = p^2/q^2$ is not true when $p,q$ have no common factors. Successive perfect squares differ by odd numbers at least 3. $\endgroup$ – Michael Oct 30 '14 at 16:54
  • $\begingroup$ @Michael Thanks for your comment! $\endgroup$ – user84413 Oct 31 '14 at 0:53
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If $k=2^r(2a+1)$ with $a\ge1$, then $\sqrt{1+{k\over n}}$ is rational when $n=2^ra^2$. If $k=2^r$ with $r\ge4$, then $\sqrt{1+{k\over n}}$ is rational when $n=2^{r-4}9$. That leaves only $k=1$, $2$, $4$, and $8$. But $k=8$ is ruled out by $n=1$. So $k=1$, $2$, and $4$ are the remaining possibilities.

Added later: Reading Nilan's answer made me realize how unnecessary it was to treat $16$ different from $8$. I was mentally locked into using the Pythagorean triple $3^2+4^2=5^2$.

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  • $\begingroup$ Thanks for your answer. Is it easy to verify that $k=1,2,4$ all satisfy this property? $\endgroup$ – user84413 Oct 30 '14 at 16:53

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