Simple algebra involving powers It is proven that $\sqrt{x^3 +1}$ is a integer with $x$ equal to some three integral values. 
I found $0$ and $2$. By trial and error. Is there a smarter way. 
 A: Yes, there are some things you can say. Call $c$ the integer you're talking about. You have $x^3 = (c-1)(c+1)$. Necessarily, the Bezout identity gives you that the gcd of $c-1$ and $c+1$ divides 2, so it is 1 or 2: $1\times(c+1) - 1\times(c-1) =2$.
If the gcd is 1, you need both $(c-1)$ and $(c+1)$ to be perfect cubes, which won't happen apart from $c+1 = 1$ and $c-1=-1$.
If the gcd is 2, $(c-1)(c+1)=4kk'$ with a gcd$(k,k')=1$ So one has to be even and the other odd, which can be rewritten $(c-1)(c+1)=8l(2m+1)$ with gcd$(l,2m+1)=1$ and both $l$ and $2m+1$ are perfect cube, so $(c-1)(c+1)=(2\alpha\beta)^3$ with gcd$(\alpha,\beta)=1$.
Any other divisor apart from 2 has to be a perfect cube and again there are no perfect cubes separated by only 2 after 0 and 1. That gives you the solution $c-1 = 2$ and $c+1 =4$.
It's a bit messy, but you have the main arguments to clarify.
A: Allowing negative numbers emits the value $-1$.
An interesting problem would be to prove that there is no perfect square that immediately follows an integer cubed other than the three values: $-1,0,2$.  I did a very quick search in Mathematica up to $x=1000$ and didn't find any others (unless I somehow messed up).
