Mathemtaical proof: n^5 - n is divisible by 10 for all n ≥2?

Question

I need some help with this particular question from the homework:

Prove: n^5 - n is divisible by 10 for all n ≥2

What steps should I take?
Thanks in advance!

Answer 1

Use induction.

Prove true for n=2
2^5=2 = 32-2 = 30 (30/10=3) So true.

Assume true for n=k
k^5 - k is divisible by 10

Then prove true for n=k+1.
(k+1)^5 - (k+1)
=(k^5+5k^4+10k^3+10k^2+5k+1) - (k+1)
=k^5+5k^4+10k^3+10k^2+4k
We know the k^5 - k is divisible by 10 (assumption) so we can subtract it, we'll need the difference to also be divisible by 10.
=5k^4+10k^3+10k^2+5k
=5k(k^3+2k^2+2k+1)
If k is even, 5k is divisible by 10, so is works for k even.
If k is odd, k^3 is odd, 2k^2 is even, 2k is even, and 1 is odd....summing 2 odds and 2 evens is an even number, so when we multiply by 5k we get a number divisible by 10, so it works for k odd.

Hope that helps.

Answer 2

n must be an integer also.

n^5 - n

Perfect example about the use of Fermat's Little Theorem.

n^5 - n

For any integer n, with this expression is divisible by 5.
Here is what the theorem states: (a^p - a) mod p = 0, if a is an integer and p is a prime number. So that's why the expression is divisible by 5.

The next step is to prove it is divisible by 2.

n^5 - n = n(n^2 + 1)(n^2 - 1)
If n is odd, n^2 + 1 and n^2 - 1 is even.
If n is even, n^2 + 1 and n^2 - 1 is odd.

So either way this expression is even and always divisible by 2.

An expression divisible by 2 and 5 is always divisible by 10. Therefore n^5 - n is divisible by 10.

Answer 3

This looks like a perfect candidate for mathematical induction.

1. ASSUME that it's true for n (where n>1)
2. SHOW that it's true for (n+1)

In step 2, you'll need to expand (n+1)^5, so you'll need to use the binomial theorem. You will be able to eliminate many of the terms because they'll have the product of 5's and 2's in them, and you should be left with something simple where you can use step 1 (ie. the assumption that it's true for n) to show that it's divisible by 10.

The final step is to show that it's true for the first value, which in this case is 2. That's easy.

Using this technique, you can show that it's true for ALL n (>1)

There IS an alternative form of this proof where step 1 is replaced by the assumption that it's true for ALL k up to n. The basic approach is the same. Sometimes this alternative form is needed when the results of step 2 are very complex and includes integers other than n and n+1.

Answer 4

n^5-n = n(n^4-1)
= n (n^2+1)(n^2-1)
= n(n^2+1)(n+1)(n-1)
Now examine the factors above and you will see that no matter what value you pick for n (n =/>2) the factors will always permit division by 10
If you pick 6 , the n-1 term times the n term gives you 30 which is divisible by 10

You should be able to work out a proof from this.

Answer 5

Use mathematical induction:

1) it's true forn=2 since 2^5-2 = 32-2 = 30

2) assume it's true for arbitrary n now prove it is true for n+1
(n+1)^5 - (n+1) =
(n^5+5*n^4+10*n^3+10*n^2+5*n+1) -(n+1) =
n^5 + 5*n^4+10*n^3+10*n^2+4*n =
(n^5 - n) + (5*n^4+10*n^3+10*n^2+5*n) =
(n^5 - n) + (5*n^4+5*n) + (10*n^3+10*n^2) =
(n^5 - n) + 5*(n^3+1)*n + (10*n^3+10*n^2)

The 1st expression in the larger expression above is divisible by 10 since we assumed it to be true as part of the induction process.

The 2nd expression in the larger expression above is divisible by 10 since it is divisible by 5 and one of (n^3+1) or n is even.

The 3rd expression in the larger expression above is divisible by 10 obviously.

So the entire expression is divisible by 10.

Therefore "if it is true for n it is true for n+1" is itself true.

Therefore it is true by mathematical induction.

Answer 6

Factoring gives n(n-1)(n+1)(n^2+1). Since one of n, n+1 has to be even, it remains to show that one of the factors has to be a multiple of 5. If n ends in 4, 5, 6, then one of n-1, n, n+1 is a multiple of 5, likewise if n ends in 9, 0, or 1. That leaves n ending in 2, 3, 7, and 8. But a number ending in 2 has a square that ends in 4, so n^2 +1 is a multiple of 5. Likewise a number ending in 8 has a square that ends in 4, and numbers that end in 3 and 7 have squares that end in 9, so n^2 + 1 is a multiple of 10 already.

Answer 7

Examining the fifth power of each digit:
1^5 = 1
2^5 = 32
3^5 = 343
4^5 = 1024
5^5 = 3215
6^5 = 7776
7^5 = 16807
8^5 = 32768
9^5 = 59049
0^5 = 0

It is obvious that n^5 - n will be a number ending in 0.

Therefore n^5 - n is divisible by 10