For what natural numbers k and n can the polynomial $x^k+x^n$ be divided by $x^2-x+1$
Please explain the solution, or tell me what is it that I should learn/look up, in order to know how to solve such problems.
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Hint: Let $L_n(x)$ be the remainder of $x^n$ on division by the polynomial $x^2 - x + 1$.
The sequence starts with
$$ \eqalign{L_0(x) &= 1\cr
L_1(x) &= x\cr
L_2(x) &= x-1\cr
L_3(x) &= -1\cr} $$
Since $L_3(x) = -L_0(x)$, we will have $L_{n+3}(x) = -L_{n}(x)$ for all $n$.
Robert Israel
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A separable polynomial (one with distinct roots) is a divisor of another polynomial if and only if every root of the first is also a root of the second. What are the roots of $x^2-x+1$? Well, this polynomial is part of the factorization of $x^3+1$, which is itself part of the factorization of $x^6-1$. Thus, the roots of $x^2-x+1$ are precisely the $6$th roots of unity.
(So, it would help to become familiar with cyclic groups, roots of unity, and cyclotomic polynomials.)
And $x^n+x^k=x^k(x^{n-k}+1)$ so you need to check of primitive $6$th roots of unity satisfy the equation $x^{n-k}=-1$. Well, what kind of powers can you take primitive $6$th roots to to get $-1$?
anon
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