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I am trying to justify why if we have a polynomial $f(x)= (x-a_1)(x-a_2)(x-a_3)$ over a prime modulus $p$,it's only roots are $a_1, a_2,$ and $a_3$. Why are there no other values of $x$ which could render this product a multiple of $p$, such that this $x$-value is a root in mod $p$? I have tried to justify that if the product of these factors for some $x$ is a multiple of $p$, then it is the case that this $x$ is congruent to one of $a_1, a_2,$ and $a_3$ but haven't got anywhere.

I am asking this solely out of curiosity, and not for a homework assignment.

Bill Dubuque
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Princess Mia
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  • Consider Euclid's lemma. – John Omielan Oct 04 '23 at 03:05
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    As here in the dupe, if $, f(a) \equiv (a!-!a_1)(a!-!s_2)(a!-!a_3)\equiv 0,$ then $p$ divides the product so by Euclid's Lemma it divides some factor $, p\mid a-a_i\ $ so $\ a\equiv a_i\ \ $ – Bill Dubuque Oct 04 '23 at 03:13
  • Suppose you plug in a value for x that is unequal to any a_j, j ∈ {1,2,3} into f(x) . Then f(x) is the product of three nonzero elements of your prime field. (I am assuming you are working over the prime field F_p, even though you did not mention this explicitly.) The product of finitely many nonzero elements in any field cannot equal zero. (This is easy to prove directly without knowing anyone else's theorem or lemma.) – Dan Asimov Oct 04 '23 at 04:14
  • @Dan OP has not yet studied ring theory (that's why I phrased my prior comment in the language of modular arithmetic). – Bill Dubuque Oct 04 '23 at 23:53
  • Sorry if what I wrote was confusing. Though F_p (p prime) might be one of the very simplest rings to study, requiring knowledge of only how to reduce a sum or product of integers mod p. (So maybe leading OP to learn about F_p might not be a bad thing.) – Dan Asimov Oct 05 '23 at 03:34

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