Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Define the map $\Phi:\mathbb{Z}/p\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$ as $\Phi(x) = x^{(p-1)/2}.$ It's easy to prove that $$\Phi(x) = 1,~~~\text{iff}~~~x = b^2~~~\text{for some}~~~b\in\mathbb{Z}/p\mathbb{Z}$$ and $\Phi(0) = 0.$ Consider the Legendre symbol $\left(\frac{n}{p}\right): \mathbb{Z}/p\mathbb{Z}\to \{\pm1, 0\}$ Then, why is it false that $\left(\frac{n}{p}\right) = \Phi(n)?$ In particular, $$\left(\frac{n}{p}\right) = 1\in\mathbb{Z},~~~\text{if}~~~\Phi(n) = 1\in\mathbb{Z}/p\mathbb{Z},$$ $$\left(\frac{n}{p}\right) = -1\in\mathbb{Z},~~~\text{if}~~~\Phi(n) = - 1\in\mathbb{Z}/p\mathbb{Z}$$ and $\left(\frac{n}{p}\right) = 0,~~~\text{if}~~~\Phi(n) = 0.$ Thanks.

share|improve this question
4  
It is not false. I don't understand what you are asking, what you write for $\Phi(n)$ can be taken to be the definition of the Legendre symbol. If you proved that $\Phi(n)$ is only 1 if $n$ is a square, I am left wondering, what is your definition of the Legendre symbol? –  Eric Naslund Jan 8 '12 at 6:06
    
Perhaps see en.wikipedia.org/wiki/Euler%27s_criterion –  ae0709 Jan 8 '12 at 6:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.