A question in to the numbertheory: give a non-trivial soloution for the equation $x^6\equiv x\ mod\ 396$. And how many soloutions does this congruence have?
I know by the Chinese Remainder Theorem that the congruence is equivalent to the system:$x^6\equiv x\ mod\ 2$, $x^6\equiv x\ mod\ 3$ and $x^6\equiv x\ mod\ 11$. Thereafter you can say that $\phi(2)=1$, $\phi(3)=2$ and $\phi(11)=10$ such that: $x^6\equiv x\ mod\ 2$, $x^3\equiv x\ mod\ 3$ and $x^6\equiv x\ mod\ 11$.
This doesn't seem to help. Can somebody help me? Thanx in advance!
$396=4\times9\times11$, so we have to solve $x^6\equiv x\pmod m$ for $m=4$, $m=9$, and $m=11$, and then use the Chinese Remainder Theorem.
By just trying everything, the solutions to $x^6=x\pmod4$ are $x\equiv0\pmod4$ and $x\equiv1\pmod4$.
Also, the solutions to $x^6\equiv x\pmod9$ are $x\equiv0\pmod9$ and $x\equiv1\pmod9$.
The solutions to $x^6\equiv x\pmod{11}$ are $x\equiv0,1,3,4,5,9\pmod{11}$.
OK?