When will $\frac{\sqrt{n^2+1}} {\sqrt{2}}$ be rational? For what n is this rational,
$$\frac{\sqrt{n^2+1}} {\sqrt{2}}$$ 
So far I have found the integers 1,7,41 and I have found some rational solutions to this as well but I'm looking to get a more general sense. 


*

*So when is this a rational number?

*Are there any restrictive properties that can be said about when this is rational? What can be said about n?


Edit: Thanks for the help on the integer solutions but is there anything that can be said about when it is rational?
 A: If $\dfrac{\sqrt{n^2+1}} {\sqrt{2}}$ is an integer $m$, then $\color{blue}{2m^2-n^2=1}$.  
This is a Pell-type equation, and there are infinitely many solutions,
including $n=1, 7, 41, 239, 1393, 8119, 47321, 275807, 1607521, 9369319$ et al.
A: Well you want to solve for $n^2+1=2y^2$, this is a standard type of Pell's Equation.
A: $\frac{\sqrt{n^2+1}}{\sqrt{2}} = \sqrt{\frac{n^2+1}{2}}$, which is rational if, and only if, $\sqrt{\frac{n^2+1}{2}}$ is an integer, so, there is $k \in \mathbb{N} : k^2 = \frac{n^2+1}{2}$
(Note that $\frac{n^2+1}{2}$ is either a reduced fraction or an integer)
If $n$ is even, there is no way it can happen, since $n^2+1$ will be an odd number and thus we won't get a perfect square
A: This number is rational in the series 1, 7, 41, 239, ...
Each new number is t(n+1)=6 t(n)-t(n-1).
The feature works because 6^2-4 is a double-square, 
EDIT:
There are a lot of solutions when it is rational, for example, putting x=23/7 will produce a rational result of 17/7.
This comes from the general pythagorus triangle $a^2-b^2 : 2ab : a^2+b^2 $.
From this, we multiply through by a rightangle (ie $1+i$), and we get
$(a^2-2ab-b^2)^2 + (a^2+2ab-b^2) = 2(a^2+b^2)^2$
This leads to the general solution:
$\frac 1{\sqrt{2}} \sqrt{(\frac{a^2+2ab-b^2}{a^2-2ab-b^2})^2+1} = \frac{a^2+b^2}{a^2-2ab-b^2}$
for all a, b that $2\mid ab$, $\gcd(a,b)=1$ and $a>b$.
