How many residue classes satisfy the congruence $x^3 \equiv 3 \pmod{21}$?
I don't understand what this question is asking me to do.
Can someone simplify the question for me, thanks.
How many residue classes satisfy the congruence $x^3 \equiv 3 \pmod{21}$?
I don't understand what this question is asking me to do.
Can someone simplify the question for me, thanks.
The question is asking you to check which of the 21 residue classes of integers modulo $21$ (to wit, the class of $0$, the class of $1$, the class of $2$, etc) are solutions to $x^3\equiv 3\pmod{21}$. If nothing else occurs to you, you can certainly plug and chug and figure out which ones are solutions and which ones are not.
Just a little something to add to Arturo's answer. Note that any solution to $x^3\equiv 3 \pmod{21}$ will also be a solution to $x^3\equiv 3\pmod{3}$ and $x^3\equiv 3\pmod{7}$. So instead of checking all $21$ congruence classes, you can begin by checking the congruence classes modulo the prime powers of $21$. If one happens to have no solution, then modulo $21$ there should be no solution either. It should save you some time, as there are fewer classes to check.
This is an application of the following theorem.
Let $f(x)$ be a fixed polynomial with integral coefficients, and for any positive interger $m$, let $N(m)$ denote the number of solutions of the congruence $f(x)\equiv 0\pmod{m}$. If $m=m_1m_2$, where $gcd(m_1,m_2)=1$, then $N(m)=N(m_1)N(m_2)$. If $m=\prod p^\alpha$ is the canonical factorization of $m$, then $N(m)=\prod N(p^\alpha)$.
HINT $\rm\ \ mod\ 7:\ \ x^3 = 3\ \Rightarrow\ x^6 = \ldots\ $ contra a well-known "little" theorem.