Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to prove $\forall m,n\in\mathbb N$:

$$ 56786730 \mid mn(m^{60}-n^{60})?$$

Thanks in advance.

share|improve this question
1  
Hint: There is a relation between the prime factors of 56786730 and the divisors of 60. –  Hagen von Eitzen Apr 2 '13 at 18:32
1  
$56786730 = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 31 \times 61$. –  Lord Soth Apr 2 '13 at 18:35
1  
Little Fermat!! –  Jyrki Lahtonen Apr 2 '13 at 18:36

2 Answers 2

up vote 7 down vote accepted

Notice that $56786730=2\times3\times5\times7\times13\times31\times 61$.

By Fermat's Little theorem, $61 \mid m^{60}-n^{60}$ as $m^{60}\equiv 1 \bmod 61$ and $n^{60}\equiv 1 \bmod 61$. [Fermat's little theorem states that $a^{p-1}\equiv 1 \bmod p$ for a prime $p$ where $\gcd(a,p)=1$]. Using casework, notice that $2 \mid mn(m^{60}-n^{60})$ (i.e. either $m$ is odd and $n$ is even or both are odd). Either way the given expression is a multiple of $2$. Notice that other factors of $56786730$ are also primes. Apply FLT again and again to finish the proof.

share|improve this answer

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

share|improve this answer
    
See this answer for a complete proof. –  Bill Dubuque Aug 10 at 23:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.