2. Find all pairs of real numbers $x$ and $y$ for which

$$
25^{x^{4}-y^{2}}+25^{y^{4}-x^{2}}=\frac{2}{\sqrt{5}}
$$
Solution. Note that

$$
\frac{25^{x^{4}-y^{2}}+25^{y^{4}-x^{2}}}{2} \geqslant \sqrt{25^{x^{4}-y^{2}} \cdot 25^{y^{4}-x^{2}}}
$$

(inequality between the arithmetic mean and the geometric mean). The right-hand side is equal to

$$
5^{x^{4}-x^{2}+y^{4}-y^{2}}=5^{\left(x^{2}-1 / 2\right)^{2}+\left(y^{2}-1 / 2\right)^{2}-1 / 2} \geqslant 5^{-1 / 2}=\frac{1}{\sqrt{5}}
$$

Thus, in the problem, we need to find $x$ and $y$ for which the inequality becomes an equality. For this, the following conditions must be satisfied:

$$
\begin{gathered}
25^{x^{4}-y^{2}}=25^{y^{4}-x^{2}} \\
\left(x^{2}-1 / 2\right)^{2}+\left(y^{2}-1 / 2\right)^{2}=0
\end{gathered}
$$

These conditions are satisfied when $x^{2}=y^{2}=1 / 2$ (and only in this case), which gives four possible solutions for $x$ and $y$.

Answer: $\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right) ;\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) ;\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right) ;\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.