It will now be shown that
i) if $A(x)= a(rx+s)^{2}(x+t) ; a,r,s,t \in \mathbb{Z}; ar\ne 0$ then there are infinite $(x,y) \in \mathbb{Z}^{2}$ so that $y^{2}=A(x)$.
ii) there is exactly 1 solution $(x,y) \in \mathbb{Z}^{2}$ so that $y^{2}=x^{2}(4x-1)$
i) $A(x) = (ar^{2}x^{2}+2arsx +as^{2}) (x+t) = ar^{2}x^{3}+2arsx^{2}+as^{2}x + ar^{2}x^{2}t + 2arsxt + as^{2}t = y^{2} $
seems to be wrong route.!
ii) computing the solutions with wolframalpha gives : $x=0, y=0$ so it is to show that there can't be any other solutions than this one. Suppose there exists $x\ne \{0\}$, then for an $x<0 \in \mathbb{Z}$ $y$ can not exist in $\mathbb{Z}$. For $x>0 \in \mathbb{Z}$ $y^{2}$ is even and thus $y$ is an even number $> 0 $
Does somebody see the right ways. Please do tell me.