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I asked the following question in the math SE, with a bounty of 200 pts, without result.

question:

To prove the quadratic reciprocity law, Gauss needed the following lemma:

If $p$ is a prime number congruent to 1 modulo 8, then there exists a prime $q<p$ such that $p$ is a non residue modulo $q$.

Gauss demonstrated this result in no 126, 127, 128, 129 of the Disquisitiones, and it is in fact the essential difficulty of the quadratic reciprocity law (with this lemma, I could provide a terrible simplification of the original proof of Gauss by induction).

I found the demonstration of Gauss both beautiful and amazing, in some sense natural too. But I feel there is something more, hidden there, a general principle or a beautiful elementary lemma of modular theory (akin to Thue's lemma for example), which would provide a much more luminous and simpler demonstration to this result.

Any insight about how to extract it from the proof of Gauss? Or maybe someone already knows how to do that? An alternative proof would be also welcome.

N.B: I'm not sure there is an online translation to English, but here is an available translation of the Disquisitiones in French (I know there is also one in German). Of course, an English translation should be available in any library.

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    $\begingroup$ I think one should not suggest (implicitly, by referring to MO as "the real professionals") of the MSE denizens that they are not professionals; plenty of professionals, including plenty of MO regulars, also hang out there. The difference is in the intended level in the questions, not necessarily in the ability of the community members. $\endgroup$
    – LSpice
    Commented Sep 19, 2021 at 3:24
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    $\begingroup$ OK, question edited to cope with your rmqs. $\endgroup$
    – MikeTeX
    Commented Sep 19, 2021 at 14:38

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In response to your second query "An alternative proof would be also welcome":

A (nearly) one-page proof of a stronger version ($q<\sqrt p$ instead of $q<p$)$^\ast$ is lemma 5.6 in Quadratic and Hermitian Forms (page 184). I made a screenshot.

$^\ast$ It seems Gauss actually proved $q<2\sqrt p$, see page 95 of the Disquisitiones.

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  • $\begingroup$ Thx. This proof is very similar to the proof of Gauss. Actually, despite it is shorter, the price is that it use more complex arguments to get more compact. +1 for the link, not really what I was looking for though. $\endgroup$
    – MikeTeX
    Commented Sep 19, 2021 at 14:48
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See L. Carlitz, "A Note on Gauss' First Proof of the Quadratic Reciprocity Theorem" Proc. Amer. Math. Soc. 11 (1960), 563-565.

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  • $\begingroup$ Nothing there about the aforementioned Lemma, but the reference the author gives, the proof of Mathew, does contain a simplification of Gauss' proof. This is still essentially the same though. $\endgroup$
    – MikeTeX
    Commented Sep 20, 2021 at 6:04

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