Hi,
How tow show that ($\displaystyle \forall n \in$ N) $\displaystyle 5n^3+n \equiv 0[6]$????
This congruence can be rewritten as:
$\displaystyle -n^3+n \equiv 0 \pmod 6$
$\displaystyle n^3 \equiv n \pmod 6$
One way is to try all possibilities: $\displaystyle n\equiv 0,\ldots,n\equiv 5$.
Another possibility is to notice that this is equivalent to two congruences
$\displaystyle n^3 \equiv n \pmod 2$
$\displaystyle n^3 \equiv n \pmod 3$
and verify the validity of them.
You can notice that one of them is exactly Fermat's little theorem for exponent 3.
I do not think that the original congruence can be obtained directly from Euler's theorem, but it might be useful for you to notice, that Euler's theorem implies this congruence for all n coprime to 6.
$\displaystyle 5n^3 + n \equiv -n^3 + n = -n(n^2-1) = -(n-1)(n)(n+1) (mod6)$
we know that the product of $\displaystyle k $ consecutive integers is the multiple of $\displaystyle k! $ so in this case , it is the multiple of $\displaystyle 3! = 6 $ , $\displaystyle -(n-1)(n)(n+1) \equiv 0 (mod6) $