The 4th Japanese Junior Mathematical Olympiad

(1) Substituting $x=1$ into the given equation, we obtain a quadratic equation $y^2+7y+10=0$, and solving this equation normally, we get $y=-2,-5$. Hence the solutions are $(x,y)=(1,-2),(1,-5)$.

(2) Assume that $(x,y)$ is a root of the given equation. Substituting the value of $x$, we obtain a quadratic equation $y^2+(x+6)y+(x^2+3x+6)=0$ of variable $y$. The roots of this equation are $$y=\frac{-(x+6)+\sqrt{(x+6{)}^2-4(x^2+3x+6)}}{2}=\frac{-(x+6)+\sqrt{-3x^2+12}}{2}$$ Since $x+6$ and $y$ are integers, $\sqrt{-3x^2+12}$ must be an integer, i.e., there exists integer $k$ such that $k^2=-3x^2+12$. One easily checks that $x$ must be $\pm 1$ or $\pm 2$ (as a square of an integer is a nonnegative integer). Solving the corresponding quadratic equation for each $x$, we conclude that the answers are $(x,y)=(-2,-2),(-1,-4),(-1,-1),(1,-5),(1,-2),(2,-4)$.