Take the 2-minute tour ×
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.

Suppose that $x$ solves $x^4-x^2+1= 0 \pmod p$. Show that $p=1 \pmod {12}$. Following a hint I have rewritten the equation as $(x^2-1)^2=-3 \mod p$ and $(2x^2-1)^2=-x^2 \pmod p$. The first equation gives that $p=1 \pmod 3$ by using quadratic reciprocity and noting that $1$ is the only applicable quadratic residue. However, I am not able to show that $p=1 \pmod 4$. Since this is a condition in some formulations of quadratic reciprocity I guess that it should be used here as well. This is exercise 13 chapter 5 from the book by Ireland and Rosen if it is any help. Thankful for any hints!

share|improve this question
3  
Maybe you could rewrite the second rewritten congruence as $\left( \frac{2x^2-1}{x}\right)^2 \equiv -1 \pmod{p}$ (clearly p cannot divide x), thereby showing that -1 is a quadratic residue. –  Daan Michiels May 6 '12 at 17:42
    
Thank you! If you post this as an answer I will gladly accept. –  user3620 May 6 '12 at 19:40

2 Answers 2

up vote 2 down vote accepted

Observe that $p$ cannot divide $x$, so that $x$ is invertible modulo $p$. Rewrite the second rewritten congruence as $$\left( \frac{2x^2-1}{x} \right)^2 \equiv -1 \pmod{p} .$$ This shows that -1 is a quadratic residue modulo $p$, implying that $p\equiv 1\pmod{4}$.

share|improve this answer

This is the minimal polynomial $f(x)$ of a primitive twelfth root of unity. It has no root mod $2$ or $3$. Therefore if $f$ has a root in $\mathbb{F}_p$ then it is a primitive twelfth root of unity since $f$ is a factor of $X^{12}-1$ and the latter has no double roots in $\mathbb{F}_p$ (the gcd of $X^{12}-1$ and $12 X^{11}$ is $1$ in $\mathbb{F}_p$). Therefore $12$ divides the order of $\mathbb{F}_p^{\ast}$. In other words $p \equiv 1 \pmod{12}$.

share|improve this answer

Your Answer

 
discard

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.