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What is the solution number of the equation

$$x^2-x+1\equiv 0\pmod{p^e}$$

I know when $e=1$, it is $1+\left(\frac{-3}{p}\right)$, and I guess it is the same for $e>1$, but can anyone provide a proof?


I know when $e=1$, the number is $$ 1+\left(\frac{-3}{p}\right) $$

When $e>1$, it is said that the answer is the same, saying that $$ 1+\left(\frac{-3}{p}\right)=1+\left(\frac{-3}{p^e}\right) $$

That's what puzzling me.

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How exactly did you obtain that result for $e = 1$? – TMM Jan 12 '13 at 15:14
@TMM, like the answer bhattacharjee is given. But I need to show $\left(\frac{-3}{p^e}\right)=\left(\frac{-3}{p}\right)$ – hxhxhx88 Jan 12 '13 at 15:29
I don't think so. We have $\left(\frac{-3}{5}\right)=-1,\left(\frac{-3}{25}\right)=-1$ – hxhxhx88 Jan 12 '13 at 15:45
Let me rephrase that: $$\left(\frac{-3}{p^e}\right)=\left(\frac{-3}{p}\right)^e.$$ So the equality is false if $e$ is even and $\left(\frac{−3}{p}\right) = -1$. In your example, $\left(\frac{−3}{25}\right) = 1$. – TMM Jan 12 '13 at 15:54
@TMM, what is the solution of $y^2\equiv-3\pmod{25}?$ – lab bhattacharjee Jan 12 '13 at 16:16
up vote 0 down vote accepted

If $p>2,$$$p^e\mid (x^2-x+1)\iff p^e\mid (2x-1)^2+3$$

So, $$(2x-1)^2\equiv-3\pmod{p^e}$$

Applying Discrete Logarithm w.r.t some primitive root $g\pmod {p^e}$,

$2ind_g(2x-1)\equiv ind_g(-3)\pmod{p^{e-1}(p-1)}$ as $\phi(p^e)=p^{e-1}(p-1)$

Using Linear congruence theorem, the last equation is solvable iff $(2,p-1)\mid ind_g(-3)\iff 2\mid ind_g(-3)$ and in that case it has exactly $(2,p-1)=2$ solutions.

Now we can prove, $-3$ is a quadratic residue of $p^e,$ iff it is a quadratic residue modulo $p$ (See below)

So, the number of solutions of $$(2x-1)^2\equiv-3\pmod{p^e}$$ is $0=1-1$ or $2=1+1$ i.e. is $1+\left(\frac{-3}p\right)$ for all prime $p>3$

[Proof: Now, if $-3$ is a quadratic residue modulo $p^s$ so there exists an integer $y$ such that $p^s\mid(y^2+3)\implies y^2+3=a\cdot p^s$ for some positive integer $a$

Now, $(y+b\cdot p^s)^2+3=y^2+3+2y\cdot b\cdot p^s+b^2p^{2s}=a\cdot p^s+2y\cdot b\cdot p^s+b^2p^{2s}$

If $p^{s+1}\mid (a\cdot p^s+2y\cdot b\cdot p^s+b^2p^{2s})\iff p\mid (a+2b)$ if $2s\ge s+1\iff s\ge 1$

$\implies 2b\equiv-a\pmod p$

so if $y^2\equiv-3\pmod{p^s}$ is solvable so will be $y^2\equiv-3\pmod{p^{s+1}}$

Using induction we say $-3$ is a quadratic residue of $p^e$ if it is a quadratic residue of $p$ for $e>1.$

Again if for $e>1,p^e\mid(y^2+3)\implies p\mid (y^2+3),$ so if $-3$ is a quadratic residue of $p^e,$ then it is a quadratic residue modulo $p$

So, $-3$ is a quadratic residue of $p^e,$ iff it is a quadratic residue modulo $p$]

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Yes, I know it. But I have to show that $\left(\frac{-3}{p^e}\right)=\left(\frac{-3}{p}\right)$, is there such conclusion? – hxhxhx88 Jan 12 '13 at 15:28
Wonderful proof! Thank you very much! – hxhxhx88 Jan 12 '13 at 15:58
@hxhxhx88, but the indices with respect to different moduli may vary, right?? – lab bhattacharjee Jan 12 '13 at 16:01
What indices?..And I'm still have a problem, why (3) is solvable iff $(2,p-1)|ind_g(-3)$? – hxhxhx88 Jan 12 '13 at 16:03
@hxhxhx88, please have a look into – lab bhattacharjee Jan 12 '13 at 16:04

There's a much easier solution: $x^2-x+1$ is a multiple of $p^e$ if and only if $(x^2-x+1)(x+1) = x^3+1$ is a multiple of $p^e$ and $x\not\equiv1\pmod p$. (Here it is important that $-1$ is not a root of $x^2-x+1$, which is true for all primes but 3.) And the congruence $x^3\equiv-1\pmod{p^e}$ means that $x^6\equiv1\pmod{p^e}$ but $x^3\not\equiv1\pmod p$, which means that $x$ has order 6 modulo $p^e$. In other words, the roots of $x^2-x+1$ modulo $p^e$ are exactly the elements of order 6 modulo $p^e$ (for $p\ne3$). Since the multiplicative group modulo $p^e$ is cyclic (for $p$ odd), the number of such elements is 2 if $6\mid (p^e-1)$ and 0 otherwise.

Similarly, the roots of the cyclotomic polynomial $\Phi_n(x)$ modulo $p^e$ are simply the elements of order $n$. The above is the case $n=6$.

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There is a typo, it should be $6\mid p^e-1$, anyway, a beautiful proof! Thank you! – hxhxhx88 Jan 13 '13 at 13:27
you're welcome! thanks for catching the typo - fixed – Greg Martin Jan 14 '13 at 4:19
3 is enough instead of $n=6$, since $-x$ has order 3 ;) – N. S. Jan 14 '13 at 5:42
Equivalently, $3\mid p^e-1$ if and only if $6\mid p^e-1$. We've reproved the fact that most primes are odd! – Greg Martin Jan 14 '13 at 21:00

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