1

Let $p$ be a prime number and $F\in\mathbb Z[x]$ a polynomial of degree less than $p-1$. Show that $$ p \mid F(0) + F(1) + \dots + F(p-1). $$

I don't really know how I should get started with the problem, I've thought about Fermat's little theorem but since $F$ is of degree less than $p-1$, it doesn't really apply. Any hints on what theorems I should consider or approach to try would be great, thanks!

user26857
  • 53,502
  • 15
  • 76
  • 167
link
  • 53
  • 1
    $$F(x) = \sum_{i=0}^{p-2}a_ix^i$$

    Consider coefficients in the sum $F(0)+\cdots + F(p-1)$. The coefficients of $a_k$ will be $$\sum_{i=0}^{p-1}i^k$$

    So, your task is the equivalent of showing that the coefficient of each $a_k$ is divisible by $p$.

     – SlipEternal Nov 18 '20 at 21:04
  • @link It's not that p divides each coefficient a_k, but it's that $p \mid \sum i^k$. – Calvin Lin Nov 18 '20 at 21:17

2 Answers2

2

Claim 1: For a fixed $ 1 \leq k \leq p-2$, $p \mid \sum_{i=1}^{p-1} i^k$.
This almost corresponds to $ F(x) = x^k$ in the question.

Proof: Work mod $p$. Let $ g$ be a primitive root modulo $p$.
Then, $ \{ 1, 2, \ldots, p-1 \} = \{ g^0, g^1, \ldots g^{p-1}\}$.
So $ \sum_{i=1}^{p-1} i^k = \sum_{j=1}^{p-1} g^{jk} = \frac{ g^{(p-1)k } - 1 } { g^ k - 1 }$ .
Note that $ p \not \mid g^k - 1$ and $ p \mid g^{(p-1)k } - 1 $, hence $p$ divides the fraction.

Claim 2: For a fixed $ 0 \leq k \leq p-2$, $p \mid \sum_{i=0}^{p-1} i^k$.
This corresponds to $ F(x) = x^k$ in the question.

Proof: For $ k = 0$, the sum is $p$, so the claim is true.
For $k>0$, this is (almost) the previous claim.

Note: For $ k = p - 1$, the statement isn't true since the sum is $ \equiv p-1 \pmod{p}$. Which of the above steps fail?

Corollary: The question is proved.
$F(x) = \sum_{k=0}^{p-2} a_k x^k,$ and the question is true for each $a_k x^k$, so it's true for the sum.

Note: We can prove Claim 1 without the machinery of primitive roots.
EG Using Method of Differences to show that for prime $n$, $ n \mid \sum_{i=1}^{n} i^k$ because the denominators of the coefficients $\mid (p-1)!$ and the numerator is a multiple of $p$.
Or appeal to Faulhaber's formula.

Calvin Lin
  • 77,906
1

Let $F(x)=\sum_{k=0}^{m}a_kx^k$ where $m\leq p-2.$ $$\sum_{i=0}^{p-1}F(i)=\sum_{i=0}^{p-1}\sum_{k=0}^{m}a_ki^k=\sum_{k=0}^{m}a_k\sum_{i=0}^{p-1}i^k=\sum_{k=0}^{m}a_k\frac{B_{k+1}(p)-B_{k+1}(0)}{k+1}=\sum_{k=0}^{m}a_k\sum_{j=0}^{k}{k+1 \choose j}B_jp^{k+1-j}$$ by Faulhaber's formula. Now, we apply another essential theorem: Von Staudt-Clausen's theorem. In the last expession above, $j\leq k\leq m <p-1$, so $p-1$ does not divide $j$. By Von-Staudt-Clausen's theorem, we conclude that $B_jp\equiv 0\pmod p$ which implies that $B_jp^{k+1-j}\equiv 0\pmod p$ for each $j,k$. Hence, $\sum_{i=0}^{p-1}F(i)\equiv 0\pmod p$.

Dr. İ Dibağ also published an article about Von Staudt-Clausen's theorem: Link.

Bob Dobbs
  • 16,473