# How several deposit courses please the harmony $x^3 \equiv 3 \pmod{21}$?

The amount of deposit courses please the harmony $x^3 \equiv 3 \pmod{21}$?

I do not recognize what this inquiry is asking me to do.
Can a person streamline the inquiry for me, many thanks.

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2019-12-02 02:48:22
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HINT $\rm\ \ mod\ 7:\ \ x^3 = 3\ \Rightarrow\ x^6 = \ldots\$ opposite a well - well-known "little" theory.

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2019-12-03 04:08:56
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The inquiry is asking you to examine which of the 21 deposit courses of integers modulo $21$ (to wit, the class of $0$, the class of $1$, the class of $2$, etc) are remedies to $x^3\equiv 3\pmod{21}$. If absolutely nothing else strikes you, you can absolutely connect and also down and also identify which ones are remedies and also which ones are not.

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2019-12-03 04:08:53
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Just a something to include in Arturo is solution. Keep in mind that any kind of remedy to $x^3\equiv 3 \pmod{21}$ will certainly additionally be a remedy to $x^3\equiv 3\pmod{3}$ and also $x^3\equiv 3\pmod{7}$. So as opposed to examining all $21$ harmony courses, you can begin by examining the harmony courses modulo the prime powers of $21$. If one takes place to have no remedy, after that modulo $21$ there needs to be no remedy either. It needs to conserve you time, as there are less courses to examine.

This is an application of the adhering to theory.

Allow $f(x)$ be a dealt with polynomial with indispensable coefficients, and also for any kind of favorable interger $m$, allow $N(m)$ represent the variety of remedies of the harmony $f(x)\equiv 0\pmod{m}$. If $m=m_1m_2$, where $gcd(m_1,m_2)=1$, after that $N(m)=N(m_1)N(m_2)$. If $m=\prod p^\alpha$ is the approved factorization of $m$, after that $N(m)=\prod N(p^\alpha)$.

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2019-12-03 04:08:04
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