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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?

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2  
$56,786,730 = 2\times 3 \times 5 \times 7 \times 11 \times 13 \times 31 \times 61$ –  J.H. Mar 15 '13 at 21:05
1  
I removed the commas between the digits, else $\ m,n\mid k\:$ may be read as $\ m\mid k\ $ and $\ n\mid k,\:$ what it usually means in number theory. –  Math Gems Mar 15 '13 at 21:58

2 Answers 2

up vote 3 down vote accepted

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

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2 case exist:

I) $\gcd(i,m)=\gcd(n,i)=1 $ that proof is complete by attention to explanation of question.

II) if one of them as example $\gcd(i,m) \not = 1$ then $\gcd(m,i)=i, i$ is prime then proof is complete by attention to $i|i\ldots$

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