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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How does one go about solving the following quadratic congruence?

$4x^2 \equiv 2 \ (\text{mod} \ 7)$

share|cite|improve this question
Gonna say the same thing I said to you last time: "In congruences, you can replace either side with that same thing plus a multiple of p." – Qiaochu Yuan Oct 29 '10 at 0:08

HINT $\ $ Multiply both sides by 2.

share|cite|improve this answer

Although not as useful in general, with a small modulus like $7$, one can let $x$ run through all possible congruence classes modulo $7$. Using Bill Dubuque's hint will make the mental calculation easier to see which $x$ actually satisfy the congruence.

share|cite|improve this answer

Since $2 \equiv 9 \pmod 7$, you have $(2x-3)(2x+3) = 4x^2-9 \equiv 0 \pmod 7$. Now use that $7$ is prime.

share|cite|improve this answer
I think there's a typo: $(2x-3)(2x+3)$. – user2468 Oct 29 '10 at 3:33
@M.S.: fixed, thanks. – lhf Oct 29 '10 at 5:11

Your Answer


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

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