# 1. CONDITION

The numbers $a$ and $b$ satisfy the equation $a^{2} b^{2} /\left(a^{4}-2 b^{4}\right)=1$. Find all possible values of the expression $\left(a^{2}-b^{2}\right) /\left(a^{2}+b^{2}\right)$.
Solution. From the given equality, it follows that $a^{4}-b^{2} a^{2}-2 b^{4}=0$, i.e., $\left(a^{2}+b^{2}\right)\left(a^{2}-2 b^{2}\right)=0$, from which $a^{2}=-b^{2}$ or $a^{2}=2 b^{2}$. The first case is impossible: the condition $a^{2}=-b^{2}$ is satisfied only by the numbers $a=b=0$, for which the given equality makes no sense. Non-zero numbers $a$ and $b$, such that $a^{2}=2 b^{2}$, satisfy the equality, and for all such $a$ and $b$ the value of the expression

$$
\left(a^{2}-b^{2}\right) /\left(a^{2}+b^{2}\right)=1 / 3
$$

Answer: $1 / 3$.