Find $x$ such that $12+13^x$ be a perfect square Find $x \in N$ such that $12+13^x$ be a perfect square
I am going to limit  $k < 12 + 13^x < k+i$ so that I can have $t<x<t+u$, I don't know how to do it, if $x=2k$, it pretty easy but x can also equal $2k +1$ too. So... Stuck here
Update 2:
I can prove that $x$ can't be $2k$, if so, x = 2k $(k \in \mathbb{N})$ then
$13^{2k}<12+13^x = 12 + 13^{2k}<(13^k+1)^2$ => $12+13^x$ can't be a perfect square.
~# if $x=2k+1$ 
=> $12+13^x = 12+13^{2k+1}$.
Now we need prove that $k$ can not greater than $1$ (how to do that ?, stuck again)

I think algebraic approach is appropriate to this problem. With some calculation, we get
$$(y+2\sqrt{3})(y-2\sqrt{3})=(4+\sqrt{3})^x (4-\sqrt{3})^x.$$
and $4\pm\sqrt{3}$ is prime on $\mathbb{Z}[\sqrt{3}]$. And
$$\frac{5-2\sqrt{3}}{4+\sqrt{3}}=2-\sqrt{3}$$
$$\frac{47+2\sqrt{3}}{(4+\sqrt{3})^3}=2-\sqrt{3}$$
So I conjectured following propositions:


*

*If $(x,y)$ is solution of this equation, then $y+2\sqrt{3}$ associates $(4+\sqrt{3})^x$ or $(4-\sqrt{3})^x$.

*And each case ($(4+\sqrt{3})^x$ associates $y+2\sqrt{3}$ or $(4-\sqrt{3})^x$ associates $y+2\sqrt{3}$) gives only one solution.
But I can't get more.
