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Let's define map $$\Phi:\mathbb{Z}/p\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$$ as $$\Phi(x) = x^{(p-1)/2}.$$ It's easy to proof that $$\Phi(x) = 1,~~~\text{iff}~~~x = b^2~~~\text{for some}~~~b\in\mathbb{Z}/p\mathbb{Z}$$ and $$\Phi(0) = 0.$$ And consider Legendre symbol $$\left(\frac{n}{p}\right): \mathbb{Z}/p\mathbb{Z}\to \{\pm1, 0\}$$ Why it's false that $$\left(\frac{n}{p}\right) = \Phi(n)?$$ More exactly, that $$\left(\frac{n}{p}\right) = 1\in\mathbb{Z},~~~\text{if}~~~\Phi(n) = 1\in\mathbb{Z}/p\mathbb{Z},$$ $$\left(\frac{n}{p}\right) = -1\in\mathbb{Z},~~~\text{if}~~~\Phi(n) = - 1\in\mathbb{Z}/p\mathbb{Z}$$ and $$\left(\frac{n}{p}\right) = 0,~~~\text{if}~~~\Phi(n) = 0.$$ Thanks.

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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 Jan 8 '12 at 6:06
Perhaps see en.wikipedia.org/wiki/Euler%27s_criterion – platinumtucan Jan 8 '12 at 6:16

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