\section*{Problem 6}

Find all integers \(x, y\) satisfying \(x^{2}+x=y^{4}+y^{3}+y^{2}+y\).


\section*{Solution}

The only solutions are \(\mathrm{x}, \mathrm{y}=-1,1-; 0,-1 ;-1,0 ; 0,0 ;-6,2 ;\) or 5,2 .

\(\left(y^{2}+y / 2-1 / 2\right)\left(y^{2}+y / 2+1 / 2\right)=y^{4}+y^{3}+1 / 4 y^{2}-1 / 4<y^{4}+y^{3}+y^{2}+y\) except for \(-1<=y<=\) \(-1 / 3\). Also \(\left(y^{2}+y / 2\right)\left(y^{2}+y / 2+1\right)=y^{4}+y^{3}+5 / 4 y^{2}+y / 2\) which is greater than \(y^{4}+y^{3}+y^{2}+\) \(\mathrm{y}\) unless \(0<=\mathrm{y}<=2\).

But no integers are greater than \(y^{2}+y / 2-1 / 2\) and less than \(y^{2}+y / 2\). So the only possible solutions have \(\mathrm{y}\) in the range -1 to 2 . Checking these 4 cases, we find the solutions listed.

