2024/051) Find integer solutions to $1+2^x+ 2^{2x+1} = y^2$

 We have 

 $1+2^x+ 2^{2x+1} = y^2$

Or  $2^x+ 2^{2x+1} = y^2-1$

Let us check one case x = 0 for which $2^x$ is odd that is 1 and we get $1 + 1+ 2 = 4 = y^2$ or 

(0, 2), (0, -2) are 2 solutions.

 $2^x(1+2^{x+1}) = y^2-1 = (y+1)(y-1)\cdots(1)$ in this x is positive

As LHS is even  number so RHS is even.

Now y+1 and y-1 both must be even and one has to be an odd multiple of power of 2 and other one odd multiple of 2.

So $y = 2^{x-1}p\pm 1$  where p is odd.

Or  $y = 2^{x-1}p +q$ where p is odd and q is $\pm 1$

Putting in the equation (1) we get

$2^x(1+2^{x+1}) = y^2-1 = (2^{x-1}p +q)^2-1 = 2^{2x-2}p^2 + 2^x pq$

Or $1+2^{x+1} = 2^{x-2} p^2 + pq$

Or  $1-pq = 2^{x-2}(p^2-8)$

q =1 gives $1-p = 2^{x-1}(p^2-8)=>(p^2-8)<0$ or $p=1$ which does not satisfy the condtion.

If q = -1 then we have

$1+p = 2^{x-2}(p^2-8)$ giving $1 + p > p^2 -8$

Or    $p^2-p -7 <=1$ this has $p  <= 3$ as p = 3 giving x = 4 and $y = \pm 23$

So 4 solutions  are $(0,2),(0,-2),(4,23), (4,-23)$