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I'm studying the theory of congruences in number theory. I found a theorem called Lagrange Theorem that state

Given a prime p , let

$f(x)=c_0+c_1x + \cdots + c_n x^n$ be a polynonomial of degree n with integer coefficents such that $ c_n \not \equiv 0 (mod p)$ then the polynoimial congruences $f(x) \equiv 0 (mod p)$ has at most n solution.

And a proof is given that is correct. Later is given an application of this Theorem in another Theorem that state:

If a polynomial f with $c_j$ coefficents is of degree n and if $f(x) \equiv 0 (mod p)$ has more that n solution where p is prime than every coefficents of f is divisibile by p.

Now the proof do not convince me or any way i could not understand. The proof is as follow If there is some coefficents not divisible by p, let $c_k$ be the one with the largest index. Then $k \leq n$ and the coungrences

$c_0 + c_1 x + \cdots + c_k x^k \equiv 0 (mod p) $

has more than k solution (Why ? this I do not understand) and so by lagrange theorem $p|c_k$ and this a contradiction.

2 Answers2

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Because since $x^k$ is the highest power of $x$ for which its coefficient is not divisible by $p$ we know that $c_{k+1}, \ldots, c_n$ are all divisible by $p$. So you have that

$$ c_0 + c_1x + \cdots + c_nx^n \equiv c_0 + c_1x + \cdots + c_kx^k \equiv 0 \ (\operatorname{mod} \ p) $$

So the congruence on the right actually has more than $n$ solutions, so in particular it has more than $k$ solutions.

Symmetry
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This holds over any field:

If a polynomial of degree at most $n$ has more than $n$ roots, then the polynomial is the zero polynomial

This follows directly from polynomial division and its corollary:

If $a$ is a root of a polynomial $f(x)$, then $f(x)=(x-a)g(x)$

Note that $\deg g = \deg f-1$. So, after $n$ roots, we get $f(x)=(x-a_1)\cdots(x-a_n)g(x)$, with $\deg g =0$.

lhf
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