4 Find all integer solutions to $x^{3}=y^{3}+2 y^{2}+1$.

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4. First note that if $y^{2}+3 y>0$, then from the original equation we deduce that $(y+1)^{3}>x^{3}>y^{3}$, which means $x^{3}$ lies between two consecutive perfect cubes, which is impossible. Therefore, it must be that $y^{2}+3 y \leqslant 0$, yielding the integer solutions $y=$ $-3,-2,-1,0$. Substituting these values into the original equation, we find all integer solutions to be $(x, y)=(1,0)$, $(1,-2),(-2,-3)$