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I'm looking at lecture notes on quadratic residues, and it states:

Let's start with the multiplicative group $Z^*_p$ (...) if $x^2 = a \mod p$ then also $(x -p)^2 = a \mod p$".

It must be a mistake that it's called a group, right? We have here both addition and multiplication...

Also, after trying to solve a question on the topic, I've encountered a situation where it would be useful if I could solve for $s$ below, given $a,t,p$, and would like to justify the following transitions, but I'm now not sure whether these are valid:

$$ a = (sp+t)^2 \mod p^2$$ $$ a = s^2p^2+2spt+t^2 \mod p^2$$ $$ a -t^2 = 2spt \mod p^2$$ $$ 2^{-1}(a -t^2)(pt)^{-1} = s \mod p^2$$

Am I right in both cases? To clarify - If I assume this is a ring, then I think I can justify the transitions up to the last one, using ring operation definitions, but since I haven't yet formally learned about rings, I'm uncertain whether that $2^{-1}$ is valid, since it came from the addition of two elements.

Thanks for any help.

Bill Dubuque
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Anon
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  • @user1090793 yes, but not every group is a ring, and they stated it's a group. – Anon Sep 13 '22 at 15:57
  • they're doing that because $\mathbb{Z}_n^*$ is $\mathbb{Z}/n\mathbb{Z} \setminus {0}$. So, it doesn't contain an additive identity, hence is not a ring with the usual operation of addition of residue classes modulo $n$. So, I should have noted in my first comment that it is incorrect to call it a ring. However, you should be able to work in $\mathbb{Z}/p^2 \mathbb{Z}$ as a ring in which case your work above is fine provided $p \neq 2$. – user1090793 Sep 13 '22 at 15:59
  • In that case, then how do we have both addition and multiplication? I recall from group theory that we had one operation. Here there are two. – Anon Sep 13 '22 at 16:04
  • See Michael's post below. He also points out (as I should have noticed, my apologies), that you need to be working in $\mathbb{Z}_p^*$ for it to be a group under multiplication. – user1090793 Sep 13 '22 at 16:06
  • Thanks, I'll sleep on it... I'm still seeing two operations: the "$-$" being the inverse of the addition operation and the "$\cdot$" between the two $(x-p)$ (i.e., the square) being the multiplication, so I can't see how this is just a group (which is always defined with a single operation "$\cdot$"). – Anon Sep 13 '22 at 16:42
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    As here, a group can be a subset of a larger set (here a field) that has additional operations. If your proof uses those additional operations then it is no longer a purely group theoretic proof but rather it lives in the larger algebraic structure. Re: your final step: $\ n x \equiv a \pmod{!n^2} \iff n\mid a\ ,&,\ x \equiv a/n \pmod{! n},,$ by congruence cancellation $\ \ $ – Bill Dubuque Sep 14 '22 at 00:47

1 Answers1

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There is no mistake in the first part. They are simply just considering the multiplicative structure on $Z^*_p$. Note also $Z^*_p$ is not closed under addition as it does not contain $0$. As for the last part it is OK down to the last line. However $(pt)^{-1}$ does not exist in $Z^*_{p^2}$.

Michael Cohen
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