\[
\left[\begin{array}{l}
\text { Derivative and multiple roots }
\end{array}\right]
\]

For which \( n \) is the polynomial \((x+1)^{n}+x^{n}+1\) divisible by:
a) \( x^{2}+x+1 \);
b) \( \left(x^{2}+x+1\right)^{2} \);
c) \( \left(x^{2}+x+1\right)^{3} \)?
Let $Q(x)=(x+1)^{n}+x^{n}+1, P(x)=x^{2}+x+1$, then $x+1 \equiv -x^{2}, x^{3} \equiv 1(\bmod P)$ (the comparison of polynomials is analogous to the comparison of numbers).

a) $Q(x) \equiv (-1)^{n} x^{2 n}+x^{n}+1(\bmod P)$. Let's consider all possible cases.

1) $n$ is a multiple of 3. Then $(-1)^{n} x^{2 n}+x^{n}+1 \equiv (-1)^{n}+2(\bmod P)$, and this polynomial is not divisible by $P$.
2) $n \equiv 1(\bmod 6)$. Then $(-1)^{n} x^{2 n}+x^{n}+1 \equiv -x^{2}+x+1 \equiv 2 x+2(\bmod P)$.
3) $n \equiv 2(\bmod 6)$. Then $(-1)^{n} x^{2 n}+x^{n}+1 \equiv x+x^{2}+1 \equiv 0(\bmod P)$.
4) $n \equiv 4(\bmod 6)$. Then $(-1)^{n} x^{2 n}+x^{n}+1 \equiv x^{2}+x+1 \equiv 0(\bmod P)$.
5) $n \equiv 5(\bmod 6)$. Then $(-1)^{n} x^{2 n}+x^{n}+1 \equiv -x+x^{2}+1 \equiv 2 x+2(\bmod P)$.

Thus, $Q$ is divisible by $P$ when $n \equiv 2,4(\bmod 6)$. In particular, $n$ is even.

b) Since the complex roots of the polynomial $P$ are distinct, it is sufficient to check whether $Q^{\prime}$ is divisible by $P$. $Q^{\prime}(x)=n(x+1)^{n-1}+n x^{n-1} \equiv n\left(x^{n-1}-(-1)^{n} x^{2 n-2}\right) \equiv n\left(x^{n-1}-x^{2 n-2}\right)(\bmod P)$.

When $n \equiv 1(\bmod 3) x^{n-1}-x^{2 n-2} \equiv 1-1 \equiv 0(\bmod P)$.

When $n \equiv 2(\bmod 3) x^{n-1}-x^{2 n-2} \equiv x-x^{2} \equiv 2 x+1(\bmod P)$.

Thus, $Q$ is divisible by $P^{2}$ when $n \equiv 4(\bmod 6)$.

c) $Q^{\prime \prime}=n(n-1)\left((x+1)^{n-2}+x^{n-2}\right) \equiv n(n-1)\left(x^{2 n-4}+x^{n-2}\right) \equiv n(n-1)\left(x+x^{2}\right) \equiv -n(n-1)(\bmod P)$. Since $n>1$, $Q$ is never divisible by $P^{3}$.

## Answer

a) $n=6 k \pm 2 ;$
b) $n=6 k-2$;
c) never.

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