Show that $y^3=4x^2-1$ has no positive integer solutions

Question

Problem: Show that $y^3=4x^2-1$ has no positive integer solutions.

I tried to consider maybe that cubes are never equal to $3 \bmod 4$, but $3^3 \equiv 3\bmod 4$.

Next I tried letting $x$ be even , so that $y$ is odd, (and if $x$ is odd, then $y$ is still odd), but that doesn't reveal much, to me at least.

Any hints appreciated.

Answer 1

This equation is satisfied if and only if some $y^3$ exists that can be expressed as a product of two consecutive odd numbers, $2x - 1$ and $2x + 1$.

Consider $p$ a prime divisor of $y$. Note that $p > 2$. One of the two factors must be divisible by $p^2$. If it's not divisible by $p^3$, then other factor must be divisible by $p$. But then both $2x - 1$ and $2x + 1$ are divisible by $p$, which means their difference $2$ must be divisible by $p$, hence $p = 2$, a contradiction.

Thus, $p^3$ must divide $2x - 1$ or $2x + 1$. This is true for all prime divisors of $y$, hence $2x - 1$ and $2x + 1$ must both be perfect cubes. There are two perfect cubes that are $2$ apart: $-1$ and $1$, thus making $x = 0$ and $y = -1$ the only solution.

Answer 2

$4x^2-1 = (2x+1)(2x-1)$

Note $\text{GCD}(2x+1,2x-1) = 1$.

If the product of two integers is a power of $n$ and their GCD is $1$, then both integers must be a power of $n$.

No two cubes differ by $2$.