Open Contests

By symmetry we can assume that $x>y$. If $x-y=1$, then $$\begin{array}{rl}\frac{x^2+4xy+y^2}{x^3-y^3}& =\frac{x^2+4xy+y^2}{(x-y)(x^2+xy+y^2)}=\frac{(x-y{)}^2+6xy}{(x-y)((x-y{)}^2+3xy)}=\\ & =\frac{1+6xy}{1+3xy}=1+\frac{3xy}{1+3xy},\end{array}$$ which is clearly not an integer. If $x-y\ge 2$, then $$\begin{array}{rl}\frac{x^2+4xy+y^2}{x^3-y^3}& =\frac{x^2+4xy+y^2}{(x-y)(x^2+xy+y^2)}\le \frac{x^2+4xy+y^2}{2(x^2+xy+y^2)}<\\ & <\frac{2x^2+2xy+2y^2}{2(x^2+xy+y^2)}=1,\end{array}$$ where the last inequality follows from $x^2-2xy+y^2=(x-y{)}^2>0$.

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Alternative solution.

If $\frac{x^2+4xy+y^2}{x^3-y^3}$ were an integer, then $$\frac{x^2+4xy+y^2}{x^3-y^3}\cdot (x-y)-1=\frac{3xy}{x^2+xy+y^2}$$

$$\begin{array}{rl}\frac{x^2+4xy+y^2}{x^3-y^3}& =\frac{x^2+4xy+y^2}{(x-y)(x^2+xy+y^2)}\le \frac{x^2+4xy+y^2}{2(x^2+xy+y^2)}<\\ & <\frac{2x^2+2xy+2y^2}{2(x^2+xy+y^2)}=1,\end{array}$$ $$wherethelastinequalityfollowsfrom$x^2 - 2xy + y^2 = (x-y)^2 > 0$.