Prove (n^5 - n) is divisible by 5 by induction

[Fizza]

here's what I've done so far...

P(n) = n^5 - n
n(n-1)(n^3+n+1)
when n = 5
5 * 4* 131 = 620
620 is a factor of 5. therefore true for n=5

assume true n=k
P(k) = k^5 - k

when n = k+1

P(k+1) = (k+1)(k+1-1)((k+1)^3 + k+2)
= (k+1)(k)(k^3 + 3k^2 + 3k + 1 + k + 2)
= (k+1)(k)(k^3 + 3k^2 + 4k + 3)
= (k^2+k)(k^3 + 3k^2 + 4k + 3)
= k^5 + 4k^4 + 7k^3 + 7k^2 + 3k

what shall I do from there?

thanks xxx

 

[mathman]

(n+1)^5=n^5+5n^4+10n^3+10n^2+5n+1
Therefore (n+1)^5-(n+1)=n^5-n+K, where K is divisible by 5.

 

[HallsofIvy]

Fizza, why do you start with n= 5 as base case? The statement is also true for n= 1, 2, 3, and 4.

Also you did not use the fact that you are assuming k⁵- k is a multiple of 5. Follow mathman's suggestion.

 

[dodo]

I'm trying to figure out how the problem was built in the first place.

I'd imagine a similar statement is true for any prime power, since the 'prime rows' of the Pascal triangle are divisible by that prime (except for the ending 1's).

 

[HallsofIvy]

So you are asking, "Is it true that n^p- n is divisible by p for p any prime?"

Yes, it is and for exactly the reason you state: if p is prime then $\left(\begin{array}{c}p \\ i\end{array}\right)$ for p prime and 0< i< p is divisible by p. That itself can be proven directly from the definition:
$$\left(\begin{array}{c}p \\ i\end{array}\right)= \frac{p!}{i!(p-i)!}$$
as long as i is neither 0 nor p, 0<p-i< p and so neither i! nor (p- i)! have a factor of p. Since p! does, the binomial coefficient is divisible by p. (We need p to be prime so that other factors in i! and (p- i)! do not "combine" to cancel p.)

Now, to show that n^p- n is divisible by p, do exactly what mathman suggested.

First, when n= 1, 1^p- 1= 0 which is divisible by p. Now assume the statement is true for some k: k^p- k= mp for some integer m. Then, by the binomial theorem,
[tex](k+1)^p= \sum_{i=0}^p \left(\begin{array}{c}p \\ i\end{array}\right) k^i[/itex]
subtracting k+1 from that does two things: first it cancels the i=0 term which is 1. Also we can combine the "k" with the i= p term which is k^p so we have k^p- k= mp. The other terms, all with 0< i< p, contain, as above, factors of p.

 

[googlymunja32]

Hello hall of Ivy

What would be the converse of this ^^^? Could you use "the assuming the opposite" method to prove whether or not this holds?

cheers

 

[LowlyPion]

here's what I've done so far...

P(n) = n^5 - n
n(n-1)(n^3+n+1)
when n = 5
5 * 4* 131 = 620
620 is a factor of 5. therefore true for n=5

Not to be too picky but your example shows n = 5 yields 620. It should be 2620, but of course satisfies the condition.

Secondly and more importantly I don't think the factors of n^5 - n are simplified enough.

I would note that:
n^5 - n = n(n^4-1) = n*(n^2-1)*(n^2+1) = (n-1)*n*(n+1)*(n^2+1)

Perhaps you can exploit the fact that for any n you necessarily have the number above and below that number that must be a factor of the result?