Let \(p\) be a prime number and let \(f(x)\) be a polynomial of degree \(d\) with integer coefficients such that:

(i) \(f(0) = 0, f(1) = 1\)

(ii) for every positive integer \(n\), the remainder upon division of \(f(n)\) by \(p\) is either 0 or 1.

Prove the \(d \geq p - 1\).

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TopNewestI am going to prove a weaker result which does not use condition (i).

Consider the polynomial \(g(x)=f(x)(f(x)-1)\) of degree \(2d\). According to (ii) \(g(x) \mod(p) = 0\) at each of \(x=0,1,2,3,\ldots, p-1\). Viewing \(g(x)\) as a polynomial over the field \(\mathbb{F}_p\), by fundamental theorem of algebra, we immediately conclude that \(2d \geq p\), i.e. \(d\geq \frac{p}{2}\). – Abhishek Sinha · 2 years, 10 months ago

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