Prove that for every prime $p$, there exists a solution to $(x^2-2)(x^2-6)(x^2-3) \equiv 0 \mod(p)$

Question

So far: I know that$\mod(p)$ is a field thus one of the factors in the expression has to equal $0$ $\mod(p)$ for some $x$. However I just don't know where to continue from this point? How do I approach a congruency like this? Are there any general tips for problems like these?

Thanks for all the help.

I have updated the title, please make sure the correction is right. — Jeel Shah, Jan 18 '14 at 15:34
Since $p$ is prime at least one of the factors is congruent to $0$. — Jay, Jan 18 '14 at 15:50

score 4 · Accepted Answer · answered Jan 18 '14 at 15:32

4

Either $\;2\;$ or $\;3\;$ is a quadratic residue modulo $\;p\;$... or else $\;2\cdot3=6\;$ is ...

answered Jan 18 '14 at 15:32

DonAntonio

211,718
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score 3 · Answer 2 · answered Jan 18 '14 at 15:32

3

Hint $\ $ mod $p,\,$ if $\,2,3$ are not squares then their product $6$ is a square, since ...

answered Jan 18 '14 at 15:32

Bill Dubuque

272,048

score 1 · Answer 3 · answered Jan 18 '14 at 15:35

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Hint: Look at the Legendre symbol.

answered Jan 18 '14 at 15:35

Harald Hanche-Olsen

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Prove that for every prime $p$, there exists a solution to $(x^2-2)(x^2-6)(x^2-3) \equiv 0 \mod(p)$

3 Answers3

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