text: # Power to the 5
  
  • Jun 7th 2012, 04:20 AM
  AndroidMachine
  Power to the 5
  Hi everyone
  
  I am having with this question:
  
  n is an integer.
  
  Show that n5 - n is divisible by 30 for all integers n.
  
  If it helps, in the questions before this one, we proved that the last digit of n5 is n.
  
  Thanks
  • Jun 7th 2012, 06:21 AM
  emakarov
  Re: Power to the 5
  Quote:
  
  Originally Posted by AndroidMachine
  If it helps, in the questions before this one, we proved that the last digit of n5 is n.
  
  So, the last digit of 7865 is 786, right? (Smile)
  
  One way is to factor n5 - n into 4 factors and then consider 5 cases depending on the remainder of n when divided by 5. In each case, find the factors among the 4 ones that are divisible by 2, 3 and 5.
  
  Another way is to use mathematical induction.
  • Jun 7th 2012, 06:27 AM
  Soroban
  Re: Power to the 5
  Hello, AndroidMachine!
  
  I'll get you started . . .
  
  Quote:
  
  $n$ is an integer.
  Show that $N \:=\:n^5- n$ is divisible by 30 for all integers $n.$
  
  We must show that $N$ is divisible by 2, 3 and 5.
  
  Factor: . $n^5 - n \;=\;n(n^4-1)$
  . . . . . . . . . . . . $=\;n(n^2-1)(n^2+1)$
  . . . . . . . . . . . . $=\;n(n-1)(n+1)(n^2+1)$
  
  $\text{We have: }\:N \:=\:\underbrace{(n-1)\cdot n\cdot(n+1)}_{\text{3 consecutive integers}}\cdot(n^2+1)$
  
  Facts: .The product of two consecutive integers is divisible by 2.
  n . . . . The product of three consecutive integers is divisible by 3.
  
  Now you must show that one of the four factors will be divisible by 5.