Primes and Functions

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 $d \geq p - 1$ .

#Algebra #Sharky

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

I 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 - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$