Legendary Product

So since Legendre symbol is multiplicative, we have (2010!/2011) and by Wilson's Theorem 2010! = -1 mod 2011, we have (-1/p) which is just -1.

For any odd prime $p$ , there exist $p - 1$ positive integers less than $p$ . Out of these $p - 1$ integers, exactly half of them, that is $\frac{p - 1 }{2}$ are quadratic residues mod $p$ , and half of them are quadratic non-residues mod $p$ . This fact is proved in the Quadratic Residues wiki .

If $1 \leq a \leq p - 1$ , then the Legendre symbol denotes:

$\left( \frac{a}{p} \right) = \begin{cases} 1 & \text{ if } a \text{ is a quadratic residue modulo } p \\ - 1 & \text{ if } a \text{ is a quadratic non-residue modulo } p \\ \end{cases}$

We saw that $\frac{p - 1 }{2}$ quadratic residues and $\frac{p - 1 }{2}$ quadratic non-residues in the interval $1 \leq a \leq p -1$ , therefore the product evaluates to

$\prod_{a = 1} ^{p - 1} \left( \frac{a}{p} \right) = (1)^{(p-1)/2} \times (-1)^{(p-1)/2}$

In this problem, $p = 2011$ , therefore

$\begin{aligned} \prod_{a = 1} ^{p - 1} \left( \frac{a}{p} \right) & = (1)^{(2011-1)/2} \times (-1)^{(2011-1)/2} \\ & = 1 \times (-1) ^{1005} \\ & = \boxed{-1} & _\square \end{aligned}$

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In general, we can say that

$\prod_{a = 1} ^{p - 1} \left( \frac{a}{p} \right) = \begin{cases}1 & \text{ if } p \equiv 1 \pmod{4}\\ -1 & \text{ if } p \equiv 3 \pmod{4} \end{cases}$

Bonus exercise: Try to prove this!