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?


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.