## Task 27/81

What values does the function $y=\sin \left(x^{8}-x^{6}-x^{4}+x^{2}\right)$ take when $x$ is an integer?
We show that the polynomial

$$
P(x)=x^{8}-x^{6}-x^{4}+x^{2}
$$

is divisible by 180 for every integer $x$. From this it follows immediately that

$$
y=\sin \left(x^{8}-x^{6}-x^{4}+x^{2}\right)=\sin \left(k \cdot 180^{\circ}\right)=0
$$

for every integer $x$ (where $k$ is also an integer). We have

$$
P(x)=x^{8}-x^{6}-x^{4}+x^{2}=[(x-1) x(x+1)]^{2}\left(x^{2}+1\right)
$$

In the square brackets, there is the product of three consecutive integers. Among them, at least one is divisible by 2 and exactly one is divisible by 3 without a remainder; consequently, the term in the square brackets is divisible by $2 \cdot 3=6$ and thus its square is divisible by 36.

Furthermore, either one of the three consecutive integers is divisible by 5 without a remainder, or it holds that $x \equiv \pm 2(\bmod 5)$ (which is easy to verify). In the first case, the square of the term in the square brackets is even divisible by $36 \cdot 25=900=5 \cdot 180$; in the second case, $x^{2} \equiv 4 \equiv-1(\bmod 5)$ and thus $x^{2}+1 \equiv 0(\bmod 5)$. Then $P(x)$ is divisible by $36 \cdot 5=180$ without a remainder (q.e.d.).