use mathematical induction to show that n^5 - n is divisible by 5 for every positive integer n.?

Question

This is a number theory question.

Answer 1

I cant figure it out because I have never taken a course in number theory or anything dealing with mathematical induction. But maybe the following proof can help. I assume you use the same method but substitute 3^2n + 2^2n-2 with n^5 - n. And if it helps n^5 - n breaks down to n(n^2 +1)(n+1)(n-1)

by mathematical induction,that the following proposition is true .
3^2n + 2^2n-2 is divisible by 5 for all natural numbers n.
When n=1,3^2+2^(2-2)=9+1=10,which is divisible by 5.
Assume n=k, 3^2k+2^(2k-2)=5M,where M is a integer.
when n=k+1
=3^[2(k + 1)] + 2^[2(k + 1) - 2]
=3^(2k + 2) + 2^(2k+2 - 2)
=(3^2k)(3^2) + 2^(2k - 2)(2^2)
=(3^2k)(9) + [ 2^(2k - 2) ](4)
=(3^2k)(4 + 5) + [2^(2k - 2)](4)
= [(3^2k)(4) + (3^2k)(5)] + [2^(2k - 2)](4)
=(3^2k)(4) + (3^2k)(5) + [ 2^(2k - 2) ](4)
Take out the common factor,4
= (4)[(3^2k) + 2^(2k-2)] + (3^2k)(5)
= (4)(5M)+ (3^2k)(5)
Take out the common factor ,5
=(5)(4M+3^2k)
so,n=k+1 is divisible by 5
By the principal of mathematical induction,P(k+1) is true of all positive integer

Answer 2

First, we need to have our base case:
if n = 1, n^5 - n = 1 - 1 = 0, which is divisible by 5.

Now let's assume that k^5 - k is divisible by 5 for some integer
k >= 1. So, k^5 - k = 5x for some integer x.

Now to do a proof by induction, we'll need to show that this statement is true for k+1, that is, we'll need to show that (k + 1)^5 - (k + 1) is divisible by 5. We can do this by writing it in the form 5*stuff, where "stuff" is an integer.

So: (k + 1)^5 - (k + 1) can be expanded using Pascal's triangle.
(for more info on this idea, see http://en.wikipedia.org/wiki/Pascal's_triangle )
We get:
k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1 - (k + 1),
the 1's cancel out, and rearranging the remaining terms gives us:
(k^5 - k) + 5k^4 + 10k^3 + 10k^2 + 5k

Notice that by our induction hypothesis, k^5 - k = 5x, so our result becomes:
5x + 5k^4 + 10k^3 + 10k^2 + 5k = 5(x + k^4 + 2k^3 + 2k^2 + k)

It should be easy for you to justify that the expression in the parentheses is an integer.

Since (k + 1)^5 - (k + 1) = 5*(an integer), it is divisible by 5. Therefore our statement is true for k+1, and by the principle of mathematical induction, it is true for all positive integers.

Answer 3

a million) for n=a million n^5-n=0 it incredibly is divisible by 5 2) think of that n^5-n is divisble by 5 and instruct that 3) (n+a million)^5-(n+a million) is divisble by 5 Making the operation =n^5 +5C1 n^4 +5C2 n^3 +5C3 n^2 5C4 n +a million -n-a million yet each and each and each and all of the words with ingredient 5C are divisible by 5 and n^5-n it incredibly is by hyp. 2) So the assumption is shown

Answer 4

1) for n=1 n^5-n=0 which is divisible by 5

2) Suppose that n^5-n is divisble by 5
and prove that

3) (n+1)^5-(n+1) is divisble by 5

Making the operation

=n^5 +5C1 n^4 +5C2 n^3 +5C3 n^2 5C4 n +1 -n-1

But all the terms with factor 5C are divisible by 5 and n^5-n it is by hyp. 2) So The theorem is proven