I'm a little out of practice when it comes to modular arithmetic, so this problem is giving me some trouble. I need to show that for every integer n, (n^3) mod 6 = n mod 6.
It is standard to write,
Note, the we can factor 6 as 2 times 3.
Where 2 and 3 are relatively prime.
Thus, it is equivalent to saying the system of congruences is satisfied:
(1)
(2)
The first congruence is true because consider cases when is even or odd. It is true in both of them.
The second congruence is true by Fermat's little theorem.
Because of the property of Zero divisors..
In general when you have a ring of multiplication mudolu , there shall always be zero divisros unless is prime.