• Nov 24th 2009, 06:51 AM
kyldn6
Prove that if c is odd, then $\displaystyle (\frac{2}{c}) = (-1)^{\frac{(c^{2}-1)}{8}}$.

I know I'm supposed to use the theorem that states $\displaystyle (\frac{2}{p}) = (-1)^{\frac{p^{2}-1}{8}}$, where p is an odd prime. But I'm not confident that stating that c is simply an odd prime is enough to satisfy this.
• Nov 24th 2009, 06:38 PM
tonio
Quote:

Originally Posted by kyldn6
Prove that if c is odd, then $\displaystyle (\frac{2}{c}) = (-1)^{\frac{(c^{2}-1)}{8}}$.

I know I'm supposed to use the theorem that states $\displaystyle (\frac{2}{p}) = (-1)^{\frac{p^{2}-1}{8}}$, where p is an odd prime. But I'm not confident that stating that c is simply an odd prime is enough to satisfy this.

I think this is false: check $\displaystyle \binom{2}{9}=-1\neq (-1)^{\frac{9^2-1}{8}}$ ...

Tonio
• Nov 24th 2009, 07:41 PM
Bruno J.
What is the symbol? The Jacobi symbol? Because the Legendre symbol is defined only for primes. If it's the Jacobi symbol, then it's true and you can prove it using the definition of the Jacobi symbol and the corresponding property of the Legendre symbol.
• Nov 24th 2009, 07:43 PM
Bruno J.
Quote:

Originally Posted by tonio
I think this is false: check $\displaystyle \binom{2}{9}=-1\neq (-1)^{\frac{9^2-1}{8}}$ ...

Tonio

$\displaystyle \left(\frac{2}{9}\right)=1$ if the symbol is Jacobi's. Note that when the Jacobi symbol $\displaystyle \left(\frac{m}{n}\right)$ is 1, it does not mean that $\displaystyle m$ is a square (mod $\displaystyle n$).
• Nov 24th 2009, 09:01 PM
kyldn6
Yes it is the Jacobi symbol.
• Nov 24th 2009, 09:39 PM
Bruno J.
Let $\displaystyle c=p_1^{\alpha_1}\hdots p_k^{\alpha_k}$ . By definition, $\displaystyle \left(\frac{2}{c}\right)=\left(\frac{2}{p_1}\right )^{\alpha_1}\hdots\left(\frac{2}{p_k}\right)^{\alp ha_k}$. Using the corresponding property of the Legendre symbol we have that this is $\displaystyle \left((-1)^{(p_1^2-1)/8}\right)^{\alpha_1}\hdots \left((-1)^{(p_k^2-1)/8}\right)^{\alpha_k} = (-1)^{(\alpha_1(p_1^2-1)+\hdots +\alpha_k(p_k^2-1))/8}$. Now what you want to show is $\displaystyle {(\alpha_1(p_1^2-1)+\hdots +\alpha_k(p_k^2-1))/8} \equiv (c^2-1)/8 \mod 2$.

Does this help a bit?
• Nov 24th 2009, 10:06 PM
kyldn6
Yes thank you so much.
• Nov 25th 2009, 08:36 PM
kyldn6
If its not too much trouble and if someone feels like it, could someone post the last half of the proof, I'm fairly confident I have it here but I'm not 100% confident.