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I am trying to understand the proof of the following result:

If $p$ is prime, and $f(x) \equiv 0 \pmod p$ with $f(x)= a_0 x^n + ... + a_n$ Then, if the congruence equation has strictly more than $n$ solutions, the coefficients of $f(x)$ are multiples of $p$.

To understand the proof I concern myself only with the case $n=2$.

So let's say there are $3$ solutions $x_1, x_2, x_3$.

We can rewrite $f(x)$ as:

$f(x) = a(x-x_1)(x-x_2) + b(x-x_1) + c$

With $a=a_0$, $b=a_1 + x_1 + x_2$ and $c = a_2 +x_1 x_2 + x_1 b$

By plugging in $x= x_1$ in $f(x)$ we get that $c\equiv 0 \pmod p$ (i.e. $p \mid c$).

However, I get in troubles when plugging in $x=x_2$ (and plugging in $x=x_3$ as well).

By plugging in by $x= x_2$, we have: $p\mid b(x_2 - x_1)$

What I want to get is that $p\mid b$, so I need to prove that $p\nmid (x_2 - x_1)$

My atempt at proving $p\nmid (x_2 - x_1)$ has been to substract $a_0 x_1 ^2 + a_1 x_1 + a_2 \equiv 0 \pmod p$ with $a_0 x_2 ^2 + a_1 x_2 + a_2 \equiv 0 \pmod p$ but the squared terms are not very easy to handle.

If anyone could help me with this I would be grateful!

Thank i.a.

niobium
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  • As in the dupes, if a quadratic has at least $2$ roots $,r_1,r_2$ then iterating the factor theorem writes it as $,f(x) = a(x-r_1)(x-r_2),$ so if $,r_3,$ is also a root then $,f(r_3)= a(r_3-r_1)(r_3-r_2)\equiv 0,,$ so $a\not\equiv 0\Rightarrow ,r_3-r_1\equiv 0,$ or $,r_3- r_2\equiv 0,,$ so every root is $\equiv r_1$ or $,r_2,,$ so there are at most $2$ roots. – Bill Dubuque Dec 18 '23 at 20:40
  • I understand that your $a$ is not congruent to $0$. But what about $b=a_1 + x_1 + x_2$ ? How can we be sure that it's not congruent to $0$ ? – niobium Dec 18 '23 at 20:40
  • Said argument proves there are at most $2$ roots. If you don't understand it then please explain what is not clear. – Bill Dubuque Dec 18 '23 at 20:43
  • I don't understand where do you use that everything is modulo a prime number, a fact that my theorem has as a premise – niobium Dec 18 '23 at 20:49
  • Is "being a domain" connected somehow to the primality of $p$ ? – niobium Dec 18 '23 at 20:55
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    $bc\equiv 0\pmod{!p}!\iff p\mid bc!!!\overset{\color{#c00}{\rm EL}!!}\iff p\mid b,$ or $,p\mid c\iff b\equiv 0,$ or $,c\equiv 0\pmod{!p},,$ by $\color{#c00}{\rm EL}$ = Euclid's Lemma and $p$ prime. Said in ring theory language: $,p,$ prime $\Rightarrow \Bbb Z_p \cong \Bbb Z/p,$ is an integral domain, i.e. $bc = 0\Rightarrow b=0,$ or $,c = 0.\ \ $ – Bill Dubuque Dec 18 '23 at 20:57
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    OP, is the contrapositive of above, i.e. if if $,r_3\not\equiv r_2$ or $,r_1$ then $,a\equiv 0,$ so all coef's of $,f,$ are $\equiv 0.\ \ $ – Bill Dubuque Dec 18 '23 at 21:05

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