Problem 5. Solve the equation: $4\left(x^{4}+3 x^{2}+1\right)\left(y^{4}-5 y^{2}+19\right)=51$.
Answer: $\left(0 ; \pm \sqrt{\frac{5}{2}}\right)$.

Solution. Note that $x^{4}+3 x^{2}+1 \geqslant 1$ for each $x$, and $y^{4}-5 y^{2}+19=\left(y^{2}-5 / 2\right)^{2}+51 / 4 \geqslant 51 / 4$ for each $y$. Therefore, the left side of the equation is greater than or equal to $4 \cdot 1 \cdot 51 / 4=51$, and equality is achieved only when $x=0$ and $y^{2}=5 / 2$. This gives the answer.