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I am trying to figure out if $3-2\sqrt{2}$ is a quadratic residue (QR) in $\mathbb{F}_p$ if $p \equiv 1 \pmod{8}$.

I know that $\sqrt{2} \in \mathbb{F}_p \iff p \equiv 1 \pmod{8}$, but I'm not sure if the proofs for that are transferable to this situation. I tried using the Legendre Symbol & the Binomial Theorem, but I didn't get very far. I also checked that the value is a QR $\forall p<100$k using Sage, so I suspect that it's true.

I am also wondering what machinery is out there for trying to figure out if some arbitrary algebraic expression is a QR (in this case the expression contains only integers & radicals) as I have another expression I want to try out: $(1+\sqrt{2})^3-(1+\sqrt{2})$

Bill Dubuque
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Stent
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    Actually, $\sqrt 2 \in \mathbb F_p \iff p\equiv \pm 1\pmod 8$. For example, $3^2\equiv 2\pmod 7$. – lulu Sep 29 '24 at 12:41
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    And what does $\sqrt 2$ mean? If $x^2\equiv 2 \pmod p$ is solvable, there are two solutions. Which one are you taking? Or are you claiming that the answer is independent of that choice? – lulu Sep 29 '24 at 12:43
  • @lulu Yes, the choice of it doesn't matter. The expression is $(a-1)^2$, so a square in any case. – Dietrich Burde Sep 29 '24 at 12:44
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    @DietrichBurde Good point. But, again, it should be $\pm 1\pmod 8$. Your argument works just as well for $-1\pmod 8$. – lulu Sep 29 '24 at 12:48
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    @DietrichBurde The choice doesn't matter in the case where you calculate $3-2\sqrt{2}$. But the OP also talks about $\sqrt{2}$ out of that context, so lulu comment stands: what does $\sqrt{2}$ mean? – jjagmath Sep 29 '24 at 12:55
  • @jjagmath It means one of the two roots of $x^2-2=0$ over the finite field $\Bbb F_p$, for $p\equiv \pm 1\bmod 8$. I don't like this notation, but it is used from time to time. – Dietrich Burde Sep 29 '24 at 13:06
  • @DietrichBurde That much is obvious, but as already pointed out, the two roots might not share all possible properties. For the question whether $3-2\sqrt{2}$ is a square the answer randomly turned out to be the same (and positive) for both choices. But as a rule of thumb, when asking a question about $\sqrt{2}$ you must either specify which of the two roots you mean, or prove that the answer does not depend on that choice. – Adayah Sep 30 '24 at 10:07

3 Answers3

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$3-2\sqrt2=(1-\sqrt2)^2$ is a square, so of course it is a quadratic residue.

Gerry Myerson
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Yes, we can apply your link. We know that $a^2=2$ for some $a\in \Bbb F_p$, for $p\equiv 1\bmod 8$. You could write $a:=\sqrt{2}$. So then we can conclude that $$ 3-2\sqrt{2}=3-2a=a^2-2a+1=(a-1)^2. $$ Hence $3-2\sqrt{2}$ is a quadratic residue modulo $p$.

Dietrich Burde
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Also, though perhaps not as elementary as one might wish as a way to address this question, there is a quadratic reciprocity law over arbitrary global fields not of characteristic $2$. A perhaps-most-memorable version is that the product of all the local quadratic Hilbert symbols is $1$, for arguments in the field: $\prod_v (a,b)_v =1$, where the Hilbert symbol is $1$ when $ax^2+by^2=z^2$ has a non-trivial solution in the $v$-th completion of the global field. From this, quadratic reciprocity follows.

This follows from more general classfield results about reciprocity laws, which are not at all trivial, but the quadratic case admits easier proofs. These are implicit (or perhaps just too-implicity stated) in A. Weil's papers about Segal-Shale-Weil repns in the early 1960s. A fairly explicit explanation is in my https://www-users.cse.umn.edu/~garrett/m/v/quad_rec_02.pdf

True, invocation of quadratic reciprocity does not instantly/algorithmically resolve all issues over general global fields, since larger class numbers certainly inhibit Euclidean algorithm parts of any approach to systematically determine whether something is a square or not mod-whatever. :)

paul garrett
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