Is $67500y^3-y$ ever a square? Is there an integer $y \neq 0$ where $67500y^3-y$ is a square?  So far I've tried a computer search with no luck.
 A: No. Let $y$ be an integer and assume $67500y^3-y = y(67500y^2-1)$ is a square. Since for any prime $p$ (EDIT: which divides $y$), the $p$-adic valuation $val_p(67500y^3-y) = val_p(y)$, $y$ has to be a positive square, or the negative of a square, or 0. In the first case, $67500y^2 -1$ must be a square too. But modulo 3, it is -1, which is not a square in $\mathbb{Z}/3$. In the second case, the product $y \cdot (67500y^2-1)$ is negative, hence not a square. Remains the case $y=0$.
A: Let's suppose that $67500y^3-y=x^2$ for some integer $x$. If we factorize we get $$y(67500y^2-1)=x^2.$$
Now, if $y<0$, the LHS is negative, but $x^2\ge 0$, so there are no solutions.
If $y>0$, since $\gcd(y, 67500y^2-1)=1$, then it's a consequence of the fundamental theorem of the arithmetic that both expressions have to be perfect squares, i.e, there are $u, v\in \Bbb{Z}$ such that $y=u^2$ and $67500y^2-1=v^2$. Looking $67500y^2-1=v^2$ mod $3$ we get $v^2\equiv 67500y^2-1\equiv -1\pmod 3$, but the perfect squares are only congruent with $0$ or $1$ mod $3$. Hence such $v$ doesn't exist, and in this case there are no solutions too. 
Therefore there is no $y\neq 0$ such that $67500y^3-y$ is a perfect square.
