B1. Find all integer solutions of the equation $m^{4}+2 n^{2}=9 m n$.
B1. If the pair $(m, n)$ is a solution to the equation, then the pair $(-m, -n)$ is also a solution, so we can assume that $m$ is non-negative. We rewrite the equation as $2 n^{2} - 9 m n + m^{4} = 0$ and consider it as a quadratic equation in $n$. Its discriminant is $81 m^{2} - 8 m^{4}$. For the quadratic equation to have integer solutions, its discriminant must be a perfect square. Therefore, $81 - 8 m^{2}$ must be a perfect square. In particular, $81 - 8 m^{2} \geq 0$, so $m \leq 3$. We can check that $81 - 8 m^{2}$ is a perfect square for $m=0$, $m=2$, and $m=3$. For $m=0$, the solution to the quadratic equation is $n=0$; for $m=2$, the solutions are $n=1$ and $n=8$; for $m=3$, the only integer solution is $n=9$. Therefore, all integer solutions to the equation are the pairs $(-3, -9)$, $(-2, -8)$, $(-2, -1)$, $(0, 0)$, $(2, 1)$, $(2, 8)$, and $(3, 9)$.[^0]