<s>[INST] why $x^3=11^3$ mod $5083$ has only one solution? Why $x^3=11^3$ mod $5083$ has only one solution? (The only answer is $x=11$)
I know it has at most 3 roots but how to find that there isn't another answer?
 [/INST] As $5083=13\cdot 17\cdot 23$
We have $\displaystyle\left(\frac x{11}\right)^3\equiv1\pmod{13}\ \ \ \ (1)$
As all primes  have primitive roots, taking discrete logarithm wrt some primitive root $g$ of $13,$(say $2$)
$3$ind$\displaystyle_g \left(\frac x{11}\right)\equiv0\pmod {12}\ \ \ \ (2)$
Using Linear Congruence Theorem,  $(2)$ hence $(1)$ have $(3,12)=3$ solutions namely, $4k\pmod{13}$ where $k=0,1,2$ of $(2)$
$\implies 2^{4k}\pmod{13},k=0,1,2$ of $(1)$
Similarly,   $\displaystyle\left(\frac x{11}\right)^3\equiv1\pmod{17}$ and $\displaystyle\left(\frac x{11}\right)^3\equiv1\pmod{23}$ should have unqiue solutions namely $1$
Using Chinese Remainder Theorem,  we shall get three in-congruent solutions $\pmod{13\cdot 17\cdot 23}$
</s>