Find the integer solution of $ a^b = 2^{2 c + 1} + 2^c + 1 $ Find the possible number of integer solution for this equation, such that $ b>1$
$$ a^b = 2^{2 c + 1} + 2^c + 1 $$
From $1$ to $1000$, $ {a = 2, b = 2, c=0} $ and $ {a = 23, b = 2, c=4} $ computationally. Are there any other possible solutions? How to show analytically that above two are it's solutions?
 A: Assume $c>2$ for simplicity's sake, so that $a$ is odd among other things.
Firstly, we note that $a^b\equiv1$ mod $2^c$. Since the order of the multiplicative group mod $2^c$ is $2^{c-1}$, we note that either $b$ is even or that $a\equiv1$ mod $2^c$, but we can rule out the latter fairly easily - $1+2*2^c$ is already too big for $b=2$, 1 is always too small and $1+2^c$ is too small for $b=2$ and too big for $b>2$.
Since $b$ is even, $a^b$ will be a square so all we need to do is show that $a=23, c=4$ is the only solution for $b=2$ and we will be done.
Rewrite the equation as:
$$(a-1)(a+1)=2^c(2^{c+1}+1)$$
$\gcd(a-1,a+1)=2$ since $a$ is odd, so it must be the case that $2^{c-1}$ divides $a-1$ or $a+1$. Let's assume $2^{c-1}|a+1$ for now. Then there exists odd numbers $u$ and $v$ such that
$$a-1=2u \\
a+1=2^{c-1}v\\
uv=2^{c+1}+1$$
Combining the first two we get $2^{c-2}v-u=1$, which we can plug into the third to obtain
$$(2^{c-2}v-1)v=2^{c+1}+1$$
We know $v$ is odd. If $v\geq5$, then the left hand side will certainly be bigger than the right hand side ($(2^{c-2}v-1)v>2^{c-3}*v^2\geq\frac{25}{16}2^{c+1}>2^{c+1}+1$) and if $v=1$, then the left hand side will definitely be smaller. So $v=3$ and
$$2^{c-2}9-3=8*2^{c-2}+1\\
2^{c-2}=4\\
c=4$$
which is exactly the solution you found. Working out the case $2^{c-1}|a-1$ is essentially the same except you find that there are no solutions then.
