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$\mathbb{Z}/{5^2\mathbb{Z}}$ ?

If we consider a polynomial $P(x) = ax^2+bx+c \equiv 0 \pmod{5^2}$ (with $\gcd(a,5^2)=1$), how many roots at most could we have for this polynomial ?

First thing is that if we consider $\mathbb{Z}/5\mathbb{Z}$ (with $\gcd(a,5)=1$) we will have at most $2$ solutions because of the field structure.

I don't think that Hensel's lemma will help in that case.

Thanks in advance !

1 Answers 1

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The usual quadratic formula works here, so we just need to know how many square roots $b^2 - 4 a c$ has in $\mathbb Z/5^2 \mathbb Z$. And that will be $5$ if $b^2 - 4 a c \equiv 0 \mod 5^2$, otherwise $0$ or $2$.

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    $4$ solutions comes from Legendre symbol ?2017-02-01
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    Sorry, typo. $5$ solutions because $x^2 \equiv 0 \mod p^2$ iff $x \equiv 0 \mod p$.2017-02-01