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For my thesis I need a good bound for the least quadratic non-residue modulo an odd prime $p$, which I can cite as proven. So I researched a lot and found several papers and bound. As far as I read in Wikipedia, there is the bound $$ p^{\frac{1}{4}\sqrt{\mathrm{e}}}. $$ This is cited by an article, but when I read the article, it says this is >>the best known estimate<<. So I get the feeling that this is not proven. Or do I understand this wrong? enter image description here

Lereu
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    "best known" means "best proved". See https://campus.lakeforest.edu/trevino/SurveyLeastNonResidue.pdf starting at Theorem 2.6 – Gerry Myerson Nov 16 '23 at 11:35
  • @GerryMyerson Thank you, I will have a look at this. I stumbled upon the word "estimate". I am German and maybe I get the term wrong, but "estimate" means, it's only guessed, not proven. Am I wrong? if yes, I would be glad :D – Lereu Nov 16 '23 at 12:26
  • An estimate can be either conjectured or proved, the word itself is neutral on this matter. You have to view it in context. $3.14159<\pi<3.1416$ are proven estimates for $\pi$. – Gerry Myerson Nov 16 '23 at 21:34
  • @GerryMyerson Theorem 5.2 in the paper you linked is exactly what I need. But I realized that it is wrong for $p = 3$, because then the bound is too low. But I want to cite this Theorem. What can I do now? – Lereu Nov 21 '23 at 17:46
  • I suppose you can cite the theorem and point out that it needs to be amended for $p=3$. Or, you could replace the $1.1$ in that theorem by a larger number that makes the statement true for $p=3$. – Gerry Myerson Nov 21 '23 at 21:20
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    @GerryMyerson Thank you. I looked up the paper by Trevino that is cited in the paper you linked. There, indeed $p>3$ is mentioned. So I used this instead. – Lereu Nov 21 '23 at 21:52

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For p = 3 or 5 (mod 8) the lowest non quadratic residue is 2. So you are looking for results where p = 1 or 7 (mod 8). This is a result of Gauss in the Disquisitiones. Theorem 5.2 you say is all you need, but p = 3 gives too low a bound. 2 is the lowest non residue possible. Thus divide primes into two segments given above and this should solve the problem.

J. H. Adams
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