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Another number theory proof

  • Thread starter ElDavidas
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Homework Statement



Let [itex]p[/itex] be an odd prime. Show that [itex] x^2 \equiv 2 (mod p)[/itex] has a solution if and only if [itex]p \equiv 1 (mod 8)[/itex] or
[itex]p \equiv -1 (mod 8) [/itex]

The Attempt at a Solution



Ok, I figured the more of these I try, the better I'll get at them. Assuming that
[itex] x^2 \equiv 2 (mod p)[/itex] has a solution first. I get

[tex] 1 = ( \frac{2}{p} )= -1^{\frac{p^2 - 1}{8}} [/tex]

So [tex] 1 = -1^{\frac{p^2 - 1}{8}} [/tex].

This implies that [itex]\frac{p^2 - 1}{8}[/itex] must be even. So

[itex] \frac{p^2 - 1}{8} = 2k[/itex] for an integer [itex]k[/itex].

I'm not sure if I'm on the right track though. Any hints would be appreciated.
 

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