Are there positive integers $a$ and $b$ such that $a(a^2-1)=2b^2$? If $a$ and $b$ are positive integers, does a solution to the follwing equation exist?
$$a(a^2-1)=2b^2$$

I have tried graphing to no avail, can anybody help?
$a(a^2-1)=a(a-1)(a+1)$. The three factors are consecutive integers and hence all coprime with the possible exception of a single factor of $2$ if $(a-1)$ and $(a+1)$ are both even. In order for three coprime factors to multiply to $2b^2$, two of the factors must be squares (the third must be twice a square). But the factors differ from each other by either $1$ or $2$, and there are no integer squares other than $0,1$ that differ from each other by either $1$ or $2$. So one solution is: $a-1=0$, $a=1$, $a+1=2$ yielding $a(a^2-1)=a(a-1)(a+1)=0$ corresponding to $b=0$. There are no other integer solutions.
$b=±\frac{\sqrt{a(a^2 -1)}}{\sqrt 2}$

You just put the value of $a$ to get the value of $b$.
And you can put any value of $a$ such that $$a(a^2-1) \ge 0$$

That is when $a \in [-1,0] \cup [1,+\infty[$