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How does one go about solving the following quadratic congruence?

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

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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

3 Answers 3

HINT $\ $ Multiply both sides by 2.

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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.

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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.

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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

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