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I am trying to find whether there is a solution to:

$$ x^2 \equiv 3\pmod {10007}$$

So I used quadratic reciprocity and found

$$ \left( \frac{3}{10007}\right) \left( \frac{10007}{3}\right) = (-1)^{\frac{10007-1}{2}}(-1)^{\frac{3-1}{2}} = (-1)\cdot (-1) = 1$$

It remains to solve $x^2 \equiv 10007 \equiv 2 \pmod 3$. It has no solutions. So that $$ \left( \frac{10007}{3}\right) = -1$$ Therefore $$ \left( \frac{3}{10007}\right) = -1$$ That would be find except I found two xolutions: $x = 1477$ or $8530$ and we verify with calculator: $$ 1477^2 - 3 = 10007 \times 218 $$ Did I do something wrong?

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Quadratic reciprocity is:

$$\left(\frac p q\right)\left( \frac q p\right)=(-1)^{\frac{p-1}{2}\cdot\frac{q-1}{2}}$$

But you've incorrectly calculated it as:

$$\left(\frac p q\right)\left( \frac q p\right)=(-1)^{\frac{p-1}{2}}(-1)^{\frac{q-1}{2}}$$

Those right sides are not equivalent.

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  • $\begingroup$ Beat me to it +1 for you, -1 for my slow error-prone typing. $\endgroup$ – Oscar Lanzi Mar 5 '17 at 2:11

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