Problem 2

For every solution with $b>1$, positivity gives $2ab^2-b^3+1=b^2(2a-b)+1>0$, hence $2a\ge b$. If $2a>b$, then $a^2>b^2$, so $a>b$. Consider the two roots $a_1,a_2$ from the preceding step. If neither satisfies $2a=b$, then, after ordering, $a_1>a_2>b$; Vieta gives $a_1+a_2=2kb^2$ and $a_1a_2=k(b^3-1)$. But then $a_1>(a_1+a_2)/2=kb^2$, so $a_1a_2>kb^3$, contradicting $a_1a_2=k(b^3-1)$. Thus one root satisfies $2a=b$. Writing $b=2t$ gives $k=t^2$ and the boundary root $(a,b)=(t,2t)$; its companion is $a'=8t^4-t$. Together with the $b=1$ family, and by direct substitution, all solutions are $(2t,1)$, $(t,2t)$, and $(8t^4-t,2t)$ for positive integers $t$.