Finding integer $x$ which its plus one becomes perfect square and its half+1 is also perfect square I am tring to find "integer $x$ which its plus one becomes perfect square and its half+1 is also perfect square". Make this statement into equation I have
\begin{align}
x + 1 = a^2, \qquad \frac{x}{2} + 1 = b^2 
\end{align}
where $x,a,b \in \mathbb{N}$. 
From knowledge of odd perfect square is of the form of $8k+1$. I noticed $x$ should be multiple of 16. 
From trial and errors I found the smallest integer $x$ is $48$.  i.e., 
\begin{align}
48 +1 = 7^2, \qquad 24+1 = 5^2
\end{align}
and the second one is 1680. What is the general form of $x$ and how one can find that? 

From the second equation, we have: $$x=2b^2-2$$
Substituing, we have:
$$2(b^2-1)-(a^2-1)=0$$
And so:
$$2b^2-a^2=1 \leftrightarrow a^2-2b^2=-1$$
This is a simply Pell equation with $d=2$. The first solution is $(a_1,b_1)\rightarrow(1,1)$ and in general: 
$$a_{n} = 6 a_{n-1} - a_{n-2}, \: \: b_{n} = 6 b_{n-1} - b_{n-2}, n\in N \land n\geq3$$
Substituing again, we have that $x$ is equal tro:
$$x=a_n^2-1=(6a_{n-1}^2-a_{n-2})^2-1$$
