Show that if $p$ is prime and $p \equiv 3 \pmod 4$ then $\frac{p-1}{2} \not\equiv \pm1 \pmod p$.


Could I say that a given is $x^2 \equiv 1 \pmod p \iff x \equiv \pm 1 \pmod p$ and then substitute in $x = \frac{p-1}{2}$ and show $(p-1) \not\equiv 1 \mod p$ giving me the final answer? If so how would I show this?

  • $\begingroup$ I can tell we will be talking about Legendre symbols soon in this Number Theory Class (I covered them last semester in Crypto I but I do not remember how to prove this). I spent 40+ min. attempting this one. $\endgroup$ – Adam Staples Mar 4 '14 at 2:04
  • $\begingroup$ math.stackexchange.com/questions/553314/… math.stackexchange.com/questions/502089/… Look similar but are a little different. $\endgroup$ – Adam Staples Mar 4 '14 at 2:08
  • $\begingroup$ The first proves more, but in particular it implies the result you are trying to prove. $\endgroup$ – André Nicolas Mar 4 '14 at 2:15
  • $\begingroup$ We haven't covered quadratic residues yet in this class and I vaguely have an idea of what they are from last semester so I'm not sure I still could use that for this proof. Someone can explain a proof involving it though if they like and I'll try and understand. $\endgroup$ – Adam Staples Mar 4 '14 at 2:18
  • $\begingroup$ I have posted an answer that does not mention QR explicitly. $\endgroup$ – André Nicolas Mar 4 '14 at 2:24

Let $q=\frac{p-1}{2}$. Note that the numbers $q+1,q+2, \dots, 2q$ are congruent modulo $p$, in reverse order, to $-1,-2,\dots,-q$. It follows that $$(p-1)!\equiv (-1)^q(q!)^2\pmod{p}.$$ But by Wilson's Theorem, we have $$(p-1)!\equiv -1\pmod{p}.$$ It follows that $$(q!)^2\equiv (-1)^{q-1}\pmod{p}.$$ If $p=4k+3$, then $q-1=2k$, which is even. Thus $$(q!)^2\equiv 1\pmod{p}.$$ The result now follows.

  • 1
    $\begingroup$ It is not, I wrote "reverse order." So for example $\frac{p+1}{2}\equiv -\frac{p-1}{2}\pmod p$. Going to the end, we have $2a=p-1\equiv -1\pmod{p}$. Do it for $p=13$. The numbers $7,8,\dots,12$ are congruent, in reverse order, to $-1,-2,\dots,-6$, or in order to $-6,-5,\dots,-1$. $\endgroup$ – André Nicolas Mar 4 '14 at 2:30
  • $\begingroup$ My bad then. ${}{}{}$ $\endgroup$ – Pedro Tamaroff Mar 4 '14 at 2:34
  • $\begingroup$ Okay yeah it's easy to see that 2($\frac{p-1}{2}$) $\equiv$ -1 (mod p) and then like you said working along 1 by 1 getting -1, -2, ... Then the rest follows. Alright thanks very much! $\endgroup$ – Adam Staples Mar 4 '14 at 2:37

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