TITLE: $\forall m,n \in \Bbb N$ : $\ 56786730\mid mn(m^{60}-n^{60})$
QUESTION [5 upvotes]: how to prove : $\forall m,n \in \Bbb N$ : $$56786730\mid mn(m^{60}-n^{60})$$
my effort:
$56786730=2.3.5.7.11.13.1841$       -Is $1841$ prime?
we must be prove: $2|m n(m^{60}-n^{60})$,...,$13|mn(m^{60}-n^{60})$ but how?
by using Fermat theorem we have if $\gcd(m,i)=1 , \gcd(n,i)=1 , (n)m^{i-1} \equiv 1 \pmod i, i=2,3,5,7,11,13$so $2|mn(m^{60}-n^{60})$,...,$13|mn(m^{60}-n^{60})$ because $i|0, i=2,3,5,7,11,13$ and $1,2,4,6,10,12 $ divide $60$ 
for other value of $m,n$ ? and factorization of $1841?

REPLY [4 votes]: Hint $\ $ Apply little Fermat, noticing that $\rm\displaystyle\ 56786730\ = \!\!\prod_{\begin{array}\rm p\ prime\\ \rm p-1\mid\, 60\end{array}}\! p$