# How many solutions to $X^6-1=0$ in $\mathbb{Z}/(504)$

How many roots of $X^6-1$ are there in $\mathbb{Z}/(504)$

I believe it's easily resolvable with the abelian groups fundamental theorem, but I want a solutions which uses only the basic notions of groups and rings.

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So you don't want it solved by decomposing $\mathbb Z/504 \cong \mathbb Z/(8) \times \mathbb Z /(9) \times \mathbb Z/(7)$? – JSchlather Jan 19 '13 at 11:53
Yes, we hadn't treated decomposition in class. Anyway, my conjecture is that the solution will be something similar to the proof of decomposition, adapted to this particular case. – Temitope.A Jan 19 '13 at 12:00
@JacobSchlather Now that you made me think about it, we have treated the decomposition you're talking about. Yes, I can prove it, it's just chines remainder theorem, isn't it? I was blided. – Temitope.A Jan 19 '13 at 12:04
Yes, it's precisely CRT. – JSchlather Jan 19 '13 at 12:15

If $n = p_1^{e_1} ... p_n^{e_n}$ with $p_i \in \mathbb{P}$ are the distinct prime factors of $n$ then $a \equiv b \mod (n)$ if and only if $a \equiv b \mod (p_i^{e_i})$ for all $i = 1, \ldots, n$. In other words, $\mathbb{Z}_n \cong \mathbb{Z}_{p_1^{e_1}} \times \ldots \times \mathbb{Z}_{p_n^{e_n}}$ via $\phi(x) = (x \mod p_1^{e_1}, \ldots, x \mod p_n^{e_n})$.

Thus, since $504 = 2^3 3^2 7$ you have $$\mathbb{Z}_{504} \cong \mathbb{Z}_{8} \times \mathbb{Z}_{9} \times \mathbb{Z}_{7}$$ So you can count the solutions in $\mathbb{Z}_{8} \times \mathbb{Z}_{9} \times \mathbb{Z}_{7}$.

$x^6 \equiv 1 \mod (8)$ has 4 solutions, $x^6 \equiv 1 \mod (9)$ has 6 solutions and $x^6 \equiv 1 \mod (7)$ has 6 solutions. Thus there are $4 \cdot 6 \cdot 6 = 144$ solutions.

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Corrected. Thank you. – fran.aubry Jan 20 '13 at 9:48