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Using this equation as an example $$5 x^2 + 7 x - 66 = 0$$ It has x=3 as a solution (also -4.4 but we'll ignore that).
Apparently we can solve this using modular arithmetic for any modulus. E.g. for modulus=4 we have $$1 \cdot x^2 + 3 x - 2 \equiv 0$$ which also yields x=3 (because $x^2$ = 9 which is congruent to 1, ditto for 3x).

I'd like to prove that this always works using ordinary algebra, but I'm not sure how. Sticking with mod 4, I tried rewriting the first equation as $$(4+1) x^2 + (4+3) x - (64+2) = 0$$ and then rearrange $$ \left[ 4 x^2 + 4 x - 64 \right] + \left[ x^2 +3 x -2 \right] = 0$$ It seems that mod 4 would allow me to remove the left bracket, but that doesn't make sense since it doesn't equal zero for x=3 (also x=3 is not a solution of the right bracket). So I'm at an impasse, any guidance would be appreciated.

Daniel
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  • It seems that your question is "If $p(x)$ is a polynomial and $n$ is an integer such that $p(n)=0$, then is $p(n)\equiv 0\pmod m$ for any modulus $m$?" If so, the answer is yes and it follows directly from the definition of congruence (and has nothing to do with polynomials). Note that it doesn't go the other way: your first congruence has $z=-1$ as a solution, but $x=-1$ is not a solution to the original equation. – Greg Martin Dec 03 '23 at 18:08
  • It is true that if $f(x)$ is a polynomial with integer coef's then any root of $f$ in $\Bbb Z$ persists as a root in $\Bbb Z_n$, by the Polynomial Congruence Rule in the first linked dupe. – Bill Dubuque Dec 03 '23 at 19:01
  • More generally roots persist mod all divisors of the modulus (above is the special case of modulus $ = 0,,$ see here), e.g. a root $!\bmod 10,$ remains a root $!\bmod 2,$ so if there are no roots $!\bmod 2,$ then there are no roots $!\bmod 10,,$ a special parity case of the modular root test. See here for more on solving modular quadratics. – Bill Dubuque Dec 03 '23 at 19:01
  • Further the above generalizes to any rational root that has denominator coprime to the modulus, e.g. see here and its links. – Bill Dubuque Dec 03 '23 at 20:27
  • Thanks for the Polynomial Congruence Rule, that's what I needed. – Daniel Dec 04 '23 at 18:57

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