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In a solution to an exercise I did they said:
For $y\in\mathbb{N}$, $y^2\equiv-1\pmod {23}$ is not possible because $23$ is a prime with $23\equiv3\pmod 4$.
Can someone explain to me this argument? I don't quite get that yet

J. W. Tanner
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Peter Mafai
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  • Have you learned about quadratic residues? – J. W. Tanner Feb 26 '24 at 10:54
  • @J. W. Tanner I just have looked at it but I am not sure how it can help here – Peter Mafai Feb 26 '24 at 11:04
  • There are rules for when $-1$ is a quadratic residue. Or do you know that $(\mathbb Z/23\mathbb Z)^\times$ is cyclic of order $22$? – J. W. Tanner Feb 26 '24 at 11:05
  • and what is this rule ? I dont know what you mean with "cyclic of order 22" – Peter Mafai Feb 26 '24 at 11:18
  • The rule is whether $-1$ is a quadratic residue mod $p$ depends on $p$ mod $4.$ Do you know there’s a primitive root mod $23$? – J. W. Tanner Feb 26 '24 at 11:36
  • If $y^2\equiv-1$ then $y^4\equiv1$ so $y^{24}/y^{22}=y^2\equiv1\pmod{23}$, contradiction – J. W. Tanner Feb 26 '24 at 11:49
  • this proof is easy to understand. but I did not use $23\equiv3(mod4)$. I would like to know how a fact about the divisor (here 23) can give information about the equation ($y^2\equiv-1(mod23)$). Before I always only used information about the other two numbers (here $y^2$,$-1$). And thats also what you did in your proof. Do you maybe know the alternative way that uses the $23\equiv3(mod4)$ fact ? – Peter Mafai Feb 26 '24 at 13:09
  • @PeterMafai the proof inherently uses the $3 \pmod 4$ condition. Let $p$ be a prime of the form $4k+3$. Then, $y^2 \equiv -1 \pmod p \implies y^4 \equiv 1 \pmod p \implies y^{4k+4} \equiv 1 \pmod p$. But, by Fermat, $y^{4k+2} \equiv 1 \pmod p$, so dividing yields $y^2 \equiv 1 \pmod p$. Find where this proof fails for $p = 4k+1$. – D S Feb 26 '24 at 13:17
  • In fact, to guarantee iff, there also exists an explicit construction for $y$ when $p = 4k+1$ using Wilson's theorem – D S Feb 26 '24 at 13:23
  • Please unaccept the answer to better facilitate duplicate processing. – Bill Dubuque Feb 26 '24 at 18:02

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