Let $f(x) = a_nx^n +a_{n−1}x^{n−1} +...+a_0$ be a polynomial of degree $n > 0$ with integer coefficients and $a_n \neq 0$. Let $p$ be a prime number and $r$ an integer. Then, if $f(r) \equiv 0 \pmod n$, there exists a polynomial $g(x)$ of degree $n − 1$ such that $(x−r)g(x)=a_nx^n +a_{n−1}x^{n−1} +...+a_1x+b_0$ where $a_0 \equiv b_0 \pmod p$.

Can anyone help me out?


closed as off-topic by Martin R, N. F. Taussig, user91500, Davide Giraudo, Sharkos Nov 10 '15 at 23:24

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I will assume that you are familiar with the Remainder Theorem. There is a polynomial $g(x)$, of degree $n-1$, with integer coefficients, and an integer constant $c$ such that $$f(x)=(x-r)g(x)+c.$$ Moreover, since $f(r)\equiv 0\pmod{p}$, we have $c\equiv 0\pmod{p}$. So we have $$a_nx^n +a_{n-1}x^{n-1}+\cdots +a_1x+a_0=(x-r)g(x)+c.$$ Now add $-c$ to both sides, and let $b_0=a_0+c$. Then $$a_nx^n +a_{n-1}x^{n-1}+\cdots +a_1x+b_0=(x-r)(g(x).$$ Moreover, since $p$ divides $c$, and $b_0=a_0-c$, we have $a_0\equiv b_0\pmod{p}$, as required.

Remark: If the aove form of the Remainder Theorem is not familiar, note that $$f(x)-f(r)=\sum a_k(x^k-r^k),$$ and $x-r$ divides $x^k-r^k$ for all $k$.


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