I am working on proving that the equation:
\begin{align} a^2-b^3 = 1 \end{align}
where $a$ and $b$ are positive integers has only one solution $(a,b) = (3,2)$.
The equation can be rewritten: \begin{align} (a-1)(a+1) = b^3. \end{align} When $b$ is odd, for any prime factor $p$ of $b$, we have that $p$ divides $(a-1)(a+1)$. But $a-1$ and $a+1$ are coprime when $b$ is odd, hence $p$ divides either $a-1$ or $a+1$ but not both. It follows that both $a-1$ and $a+1$ must be perfect cubes. Since there are no pair of perfect cubes with difference 2, it follows that $b$ must be even.
In this case let $b = 2n$ and for ease of reading, let $a = m+1$. The equation can be rewritten: \begin{align} m(m+2) = 8n^3. \end{align} This implies that $m$ is even, i.e. $m=2k$. The equation becomes: \begin{align} k(k+1) = 2n^3. \end{align} Depending on wether $k$ or $k+1$ is even, we have: \begin{align} q(2q+1) = n^3 \end{align} when $k=2q$, and \begin{align} (q+1)(2q+1) = n^3 \end{align} when $k=2q+1$.
When $k = 2q$, since $\text{gcd}(q,2q+1) = 1$, we have that $q$ and $2q+1$ must be perfect cubes using the same reasoning as above. I would like to conclude by proving that there is no positive integer $c$ such that $2c^3+1$ is a perfect cube, hence the question in the title.
