Problem
NT-B2-M12-P017 When \(n^2\mid f(n)\)
#17
★★★★☆ Level 4 of 5
Let \(f\in\mathbb Z[x]\). Prove that \(n^2\mid f(n)\) for all positive \(n\) if and only if \(f(x)=x^2g(x)\) for some \(g\in\mathbb Z[x]\).
First get \(x\mid f(x)\) from \(n\mid f(n)\), then apply the same argument to the quotient.
If \(n^2\mid f(n)\), then in particular \(n\mid f(n)\), so by the criterion for \(n\mid f(n)\), \(f(0)=0\), hence \(f(x)=xh(x)\) with \(h\in\mathbb Z[x]\). Now \(n^2\mid f(n)=nh(n)\) gives \(n\mid h(n)\) for all \(n\). Applying the criterion again, \(h(0)=0\), so \(h(x)=xg(x)\). Therefore \(f(x)=x^2g(x)\). The converse is immediate: \(f(n)=n^2g(n)\).
A good problem on applying the same tool twice.