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

Edit:

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?

Rócherz
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Adam Staples
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  • 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. – Adam Staples Mar 04 '14 at 02:04
  • http://math.stackexchange.com/questions/553314/prove-that-left-fracp-12-right-equiv-1n-mod-p-n-is-quad-no?rq=1

    http://math.stackexchange.com/questions/502089/prove-that-ap-1-2-equiv-1-mod-p-and-ap-1-2-equiv-1-mod-p?rq=1

    Look similar but are a little different.

     – Adam Staples Mar 04 '14 at 02:08
  • The first proves more, but in particular it implies the result you are trying to prove. – André Nicolas Mar 04 '14 at 02:15
  • 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. – Adam Staples Mar 04 '14 at 02:18
  • I have posted an answer that does not mention QR explicitly. – André Nicolas Mar 04 '14 at 02:24

1 Answers1

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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.

André Nicolas
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    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 $-6,-5,\dots,-1$. – André Nicolas Mar 04 '14 at 02:30
  • My bad then. ${}{}{}$ – Pedro Mar 04 '14 at 02:34
  • 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! – Adam Staples Mar 04 '14 at 02:37