Task B-1.5. Determine all integers $x, y$ for which $y^{4}+x^{2010}=2 y^{2}-1$.
Solution.

$y^{4}+x^{2010}=2 y^{2}-1$

$y^{4}-2 y^{2}+1+x^{2010}=0$

$\left(y^{2}-1\right)^{2}+x^{2010}=0$

(1 point)

The square of any real number is non-negative, i.e., $\left(y^{2}-1\right)^{2} \geq 0$ and $x^{2010} \geq 0$.

We conclude that $\left(y^{2}-1\right)^{2}=0$ and $x^{2010}=0$, which means $y_{1,2}= \pm 1$ and $x=0$.

(1 point)

The solutions are: $(0,-1)$ and $(0,1)$.

(1 point)