Can an even square exceed a cube by one? Can an even square exceed a cube by one?
I can't find any examples so I assume it is false. How would I prove it?
I forgot to add that they must be positive.
 A: Here is a proof using elementary methods:
We want to find positive integer solutions to the equation $(2m)^2 = n^3 + 1$. This is equivalent to $$ n^3 = (2m+1)(2m-1).$$
If $p$ is any prime in the prime factorization of $2m+1$ then from the equation above $p$ divides $n^3$. Because $n^3$ is a perfect cube, the exponent of $p$ in the prime factorization of $n^3$ must be divisible by 3. Then since $2m+1$ and $2m-1$ are relatively prime, $2m+1$ must contain all of $n^3$'s factors of $p$. So the exponent of $p$ in the prime factorization of $2m+1$ must be divisible by 3. This is true for every prime factor $p$ of $2m+1$, so $2m+1$ is a perfect cube. 
Using a similar argument we see that $2m-1$ must also be a perfect cube. 
So we have two perfect cubes whose difference is $(2m+1)-(2m-1) = 2$. But this is impossible, since the difference between two (distinct) positive perfect cubes is at least $2^3 -  1^3 = 7$. 
Hence the equation $(2m)^2 = n^3 + 1$ has no positive integer solutions.
A: If I remember correctly, the elliptic curve $y^2=x^3+1$ only has $(0,\pm 1)$, $(-1, 0)$ and $(2, \pm 3)$ as rational points (not including infinity). The only solution with $y$ even was already mentioned in the comments. 
Edit': Thanks to Álvaro Lozano-Robledo for finding the reference, and Keith Conrad for informing me that the result is only quoted in the reference, not actually proven. Nonetheless, I am now sure that the result is true, and have an idea of the proof.  Álvaro Lozano-Robledo suggested a strategy using $2$-descent to show that the rank of this curve is $0$, and then using the Nagell-Lutz theorem to find the points of finite order. I will return to this post in some finite amount of time after doing some more reading and see if I can supply the proof.
A: You are looking for positive integer solutions to $(2a)^2 = b^3+1$, or equivalently for $(2a-1)(2a+1)=b^3$.
Now we have two consecutive odd numbers on the LHS, so they are relatively prime (any common factor has to divide $2$, and they are odd).  Hence they both must be cubes.
Now it is easy to show that two consecutive cubes $x^3$ and $(x+1)^3$ are separated by $3x^2+3x+1$ which would be more than $2$ for positive $x$.
A: Catalan's conjecture states that  there are no two powers (with exponents $\geq2$) of natural numbers  at distance $1$ from each other, apart from $8$ and $9$. The conjecture has been proven by Preda Mihailescu in 2002.
