# 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: $$x=a_n^2-1=(6a_{n-1}^2-a_{n-2})^2-1$$

We have: $$2b^2-1=(x+2)-1=x+1=a^2 \implies a^2-2b^2=-1$$

This is a Pell Equation. We can observe that: $$(a_1^2-2b_1^2)(a_2^2-2b_2^2)=(a_1a_2+2b_1b_2)^2-2(a_1b_2+a_2b_1)^2=A^2-2B^2$$

This means that multiplying values of the form $$x^2-2y^2$$ will result in values of the same form. It is easy to see that the first solution is $$(a,b)=(1,1)$$. Now, we can see that since $$a^2-2b^2=1$$, the expression $$(a^2-2b^2)^{2k-1}=-1$$ will generate solutions when $$k \in \mathbb{N}$$. For example,

$$k=2 \implies (1^2-2\cdot1^2)^3=(3^2-2\cdot2^2)(1-2\cdot1^2)=7^2-2\cdot5^2=-1$$

This gives the solution $$(a,b)=(7,5)$$ that you had gotten.

$$k=3 \implies (1^2-2\cdot1^2)^5=(7^2-2\cdot5^2)(3^2-2\cdot2^2)=41^2-2\cdot29^2=-1$$

This gives $$(a,b)=(41,29)$$ and so on...