Find the values of $k$ for which $\sqrt{1+\frac{k}{n}}$ is irrational for all $n\in\mathbb N$

Question

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 $k$.

Answer 1

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 only need to consider powers of two. 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 remained the cases $$k=1,2,4$$ which can be treated individually.

Answer 2

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