Let $a,b \in \mathbb{N}$. How to show that there is no other solution to $$ \frac{\left(\frac{3}{2}\right)^b - 1}{2^a - \left(\frac{3}{2}\right)^b} \in \mathbb{N} $$ other than $a = b = 1$ ?
My attempts so far:
I solved for $\left(\frac{3}{2}\right)^b$:
$$ \left(\frac{3}{2}\right)^b = \frac{2^ac + 1}{c + 1} $$
Then I equaled the numerator and the denominator such that:
$$ \begin{align} 3^b &= 2^ac+1 \\ 2^b &= c + 1 \end{align} $$
which is forming combined the equation:
$$ 3^b = 2^{a+b} - 2^a + 1 $$
Then I looked for the binary representations of these numbers. The right hand side is regulary splitted in three regions. Some 1 at the beginning, some zeros following and a single 1 at the end. But now I stuck because I cant find asimilar regularity in $3^b$.
Rewrite your last equations as: $$ 3^b - 1 = 2^a(2^b-1)$$ Assume that $a>1$. Then the RHS is divisible by $4$. Note that when $b$ is odd the LHS is $2$ modulo $4$ which is a contradiction. So $b$ must be even, eg $b = 2x$. Then $2^b-1 = 4^x-1 = 0 \mod 3$. This means that the RHS is divisible by $3$ but the LHS clearly isn't. So we have that $a$ must be equal to $1$.
Can you finish from here?