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Composite Zeros - Problem 8

Let $$f(x)$$ be a polynomial over the rationals. Show that there exists a nonzero polynomial $$g(x)$$ over the rationals such that the coefficient of $$x^k$$ in $$f(x)g(x)$$ is $$0$$ for $$k=0,1$$, and all composite $$k\ge2$$.

Note by Cody Johnson
3 years, 11 months ago

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Let $$n$$ be the degree of $$f(x)$$. We have to show that $$f(x)$$ is a divisor of a polynomial of the form $$\sum q_k x^{p_k}$$, where all the $$q_k$$s are rational numbers and all the $$p_k$$s are prime numbers. Let $$p_1,\ldots,p_{n}$$ be the first $$n$$ prime numbers and $$g_j(x) = x^{p_j}\pmod{f(x)}$$ for $$j=1,\ldots,n$$. Since the degree of every $$g_j\in\mathbb{Q}[x]$$ is at most $$n-1$$ and there are $$n$$ of them, there exists a linear combination of $$g_1,\ldots,g_n$$ such that $q_1\cdot g_1(x) + \ldots + q_n\cdot g_n(x)\equiv 0$ and every $$q_i$$ is a rational number. This gives that $$f(x)$$ divides $$q_1\cdot x^{p_1}+\ldots+q_n\cdot x^{p_n}$$, as wanted.

- 3 years, 11 months ago

Jack do you compete in Proofathon? You would kick butt. :D

- 3 years, 8 months ago