I am attempting to solve the following problem:
Prove that $2^{2a+1}+2^a+1$ is not a perfect square for every integer $a\ge5$.
I found that the expression is a perfect square for $a=0$ and $4$. But until now I cannot coherently prove that there are no other values of $a$ such that the expression is a perfect square.
Any help would be very much apreciated.
I will assume that $a \ge 1$ and show that the only solution to $2^{2a+1}+2^a+1 = n^2$ is $a=4, n=23$.
This is very non-elegant but I think that it is correct. I just kept charging forward, hoping that the cases would terminate. Fortunately, it seems that they have.
If $2^{2a+1}+2^a+1 = n^2$, then $2^{2a+1}+2^a = n^2-1$ or $2^a(2^{a+1}+1) = (n+1)(n-1)$.
$n$ must be odd, so let $n = 2^uv+1$ where $v $ is odd and $u \ge 1$.
Then $2^a(2^{a+1}+1) = (2^uv+1+1)(2^uv+1-1) = 2^u v(2^u v+2) = 2^{u+1} v(2^{u-1} v+1) $.
If $u \ge 2$, then $a = u+1$ and $2^{a+1}+1 =v(2^{u-1} v+1) $ or $2^{u+2}+1 =v(2^{u-1} v+1) =v^22^{u-1} +v $.
If $v \ge 3$, the right side is too large, so $v = 1$. But this can not hold, so $u = 1$.
Therefore $2^a(2^{a+1}+1) = 2^{2} v( v+1) $ so that $a \ge 3$.
Let $v = 2^rs-1$ where $s$ is odd and $r \ge 1$. Then $2^{a-2}(2^{a+1}+1) = v( 2^rs) $ so $a-2 = r$ and $2^{a+1}+1 = vs \implies 2^{r+3}+1 = vs = (2^rs-1)s = 2^rs^2-s $.
Therefore $s+1 =2^rs^2-2^{r+3} =2^r(s^2-8) \ge 2(s^2-8) \implies 2s^2-s \le 17$ so $s = 1$ or $3$.
If $s = 1$, then $2^{r+3}+1 =2^r-1 $ which can not be.
If $s = 3$ then $2^{r+3}+1 =9\cdot 2^r-3 \implies 4 =9\cdot 2^r-2^{r+3} =2^r \implies r = 2, v = 11, a = 4$ and $2^9+2^4+1 =512+16+1 =529 =23^2 $.