Prove or disprove that $\sqrt{2+\frac{1}{n}}$ is irrational for $n \in \mathbb{Z}^+$ I have good reason to suspect that $\sqrt{2+\frac{1}{n}}$ is irrational for all $n \in \mathbb{Z}^+$ but a proof of this eludes me. I've tried proof by contradiction have had no success. I've also tried induction where clearly the base case is true, i.e. $\sqrt{2+1}=\sqrt{3}$ is irrational, but I haven't been able to show the induction step.
 A: In fact, there are infinitely many $n$ with $\sqrt{2+\frac1n}$ rational. 
You need $n$ and $2n+1$ to both be perfect squares, say $y^2$ and $x^2$. Then you are trying to solve:
$$x^2-2y^2=1\tag{1}.$$
Then $n=y^2$.
There are infinitely many solutions to this equation, starting with $(x_1,y_1)=(3,2)$ and $(x_{k+1},y_{k+1})=(3x_k+4y_k,2x_k+3y_k)$. So the first is $n=4$, next $n=144$. Next: $n=4900$.
You can actually get the closed form:
$$n =\left\lfloor\frac{(17+12\sqrt 2)^{k}}{8}\right\rfloor.$$
The equation (1) is called Pell's Equation for $2$.
Indeed, if $D$ is a positive integer, then:


*

*If $D$ is not s perfect square then there are infinitely many positive $n$ such that $\sqrt{D+\frac1 n}$ is rational;

*If $D$ is a perfect square, then there is no positive $n$ such that $\sqrt{D+\frac1n}$


This all follows from elementary results about Pell's equation. 
A: Disproof: $n=4 \rightarrow 2+1/4= 2.25 ; \sqrt{2.25} =1.5=3/2$
A: Let's start a proof by contradiction to try to show that no solutions exist and see where the proof goes wrong.  Suppose that $\sqrt{2+\frac{1}{n}}$ is rational and can be written as $\frac{p}{q}$ where $p$ and $q$ share no factors.  Then, we know that
$$
2+\frac{1}{n}=\frac{p^2}{q^2}.
$$
By clearing $q^2$ from both sides, we have that 
$$
2q^2+\frac{q^2}{n}=p^2.
$$
Therefore, $\frac{q^2}{n}$ must be an integer, so $n\mid q^2$.  
Now, if $\frac{q^2}{n}\not=1$, then there is some prime $s$ dividing $\frac{q^2}{n}$.  Therefore, this prime divides $q$.  Then, it follows that this prime divides the LHS so it divides $p^2$, but then $p$ and $q$ share a factor (a contradiction).
Therefore, it must be that $q^2=n$.  Therefore, the equality becomes:
$$
2n+1=p^2
$$
(which is the same as $2q^2+1=p^2$).
We now begin the construction:  Let us assume that $n$ is a square integer such that $2n+1$ is a square.  Let $p=\sqrt{2n+1}$ and $q=\sqrt{n}$.  Then $\sqrt{2+\frac{1}{n}}=\frac{p}{q}$ and this is the only way to have a counterexample (as seen in the provided disproof).
