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 \(k\ge2\).

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TopNewestLet \(n\) 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. – Jack D'Aurizio · 3 years, 3 months ago

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– Finn Hulse · 3 years ago

Jack do you compete in Proofathon? You would kick butt. :DLog in to reply