Question 498334: Show that for all positive integers n, (n^5)-n is divisible by 5
Answer by tinbar(133)   (Show Source):
You can put this solution on YOUR website! You can prove this using induction. The basic idea behind induction is that you first show for at least ONE actual value of n(usually n =1), your equation is true. Putting that aside for a minute convince yourself that if the equation holds true for SOME value, and that further if you have two successive values and a way to show that if the first one is true, then the second one is true (ie if n works, then n+1 works) then it MUST be true that all values after your 'base case'(n=1) is proven to be true along with a way to prove if n is true then n+1 is true.
Now for the proof:
1) Let's prove the base case: n=1 means n^5-n = 1-1 = 0, and yes it's true that 0 is divisible by 5, 5*0=0. This particular equation gives a very trivial value so if you're still unsure about it's validity, try some more values of n.
2) Let's assume for some n, the equation n^5-n is divisible by 5. Is (n+1)^5-(n+1) divisible by 5? Let's expand all of (n+1)^5 -(n+1). I'm going to skip the algebra, but it will be up to you to show your work, otherwise the answer is not complete. If you are unsure on how to expand this look up http://en.wikipedia.org/wiki/Binomial_theorem
...After expansion we get (n^5+5n^4+10n^3+10n^2+5n+1)-(n+1). The trick with induction at this point is to try and separate the equation for 'n' not 'n+1.' So the equation for n is n^5-n. Let's try to find this in the equation for 'n+1.' From the first bracket, the first term is n^5 and from the second bracket, with the negative sign in front, the first term is -n. So what's left behind? We have 5n^4+10n^3+10n^2+5n+1-1 = 5n^4+10n^3+10n^2+5n. So, basically we can simplify and regroup (n+1)^5-(n+1) into (n^5-n)+(5n^4+10n^3+10n^2+5n). Clearly the first bracket is divisible by 5, we took it as assumption that this n works. The second bracket has a factor of 5 in each term, so by definition it can be divided by 5. So in essence, we've shown if some n is true, then by 'connecting' n to the next term n+1, we have that n+1 true as well.
3) We now know 1 works, and we know the number after 1 must work, so 2 is true. Then since we know 2 works, the number after 2 must work, so 3 is true. This obviously goes on forever and so we have shown all n must work.
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