Timeline for Does there exist a non-square number which is the quadratic residue of every prime?

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Jul 7, 2013 at 18:30	history	edited	brando	CC BY-SA 3.0	edited body
Jul 7, 2013 at 18:30	comment	added	brando		Ah right, what I had in my head was $(q\|p)=-1$, not $q=-1\mod p$.
Jul 5, 2013 at 22:36	comment	added	Michael Zieve		In the last step you say $(q\|p)^n=-1$, which is only true when $(q\|p)=-1$. Since $q\equiv -1\pmod{p}$, this means that $(-1\|p)=-1$, so $p\equiv 3\pmod{4}$. Although now that I write this, I see that you could have replaced your condition $q\equiv -1\pmod{p}$ with the condition that $q$ is a nonsquare mod $p$, and then your proof would work whenever $a$ has an odd prime factor. And of course for the remaining values of $a$ one can explicitly describe a prime modulus in which $a$ is a nonsquare.
Jul 5, 2013 at 13:06	comment	added	brando		Where is the assumption? I can't see it. Thanks.
Jul 5, 2013 at 4:46	review	First posts
Jul 5, 2013 at 5:35
Jul 5, 2013 at 4:39	comment	added	Michael Zieve		In the last step you implicitly assumed that $p\equiv 3\pmod{4}$. So your argument proves the result for non-squares which have a prime factor congruent to $3\pmod{4}$.
Jul 5, 2013 at 4:31	history	answered	brando	CC BY-SA 3.0