prove that 30 I (n^5-n) using math theory?

Answer 1

I answered this same question a week or two ago.

n^5 - n
= n(n^4 - 1)
= n(n² - 1)(n² + 1)
= n(n - 1)(n + 1)(n² + 1)

The first three terms n, n - 1, and n + 1 represent three consecutive integers. So at least one of those integers must be divisible by 3. Also, at least one of them must be even. So n^5 - n must be divisible by 6...now we just have to find a way to show that at least one of the factors is divisible by 5.

If any of n, n - 1, or n + 1 is divisible by 5, then the product must be divisible by 5, and therefore n^5 - n is also divisible by 6 x 5 = 30, from the last paragraph.

If none of n, n - 1, and n + 1 is divisible by 5, then n mod 5 must be 2 or 3. (This means that, when divided by 5, n must have a remainder of 2 or 3. So, for example, if n were 7 (7 mod 5 = 2), n - 1 is 6 and n + 1 is 8, and none of them is divisible by 5. For another example, if n were 13 (13 mod 5 = 3), n - 1 is 12 and n + 1 is 14, and none of these is divisible by 5.)

In the case of n mod 5 = 2, we can write n as 5k + 2. Then n² + 1 = (5k + 2)² + 1 = 25k² + 20k + 4 + 1 = 25k² + 20k + 5 = 5(5k² + 4k + 1), which is clearly a multiple of 5.

In the case of n mod 5 = 3, we can write n as 5k + 3. Then n² + 1 = (5k + 3)² + 1 = 25k² + 30k + 9 + 1= 25k² + 30k + 10 = 5(5k² + 6k + 2), which again is clearly a multiple of 5.

Therefore, no matter what, n^5 - n is divisible by 30.

Answer 2

You can also try mathematical induction.

First, n must be at least 2.

n = 2, (n^5-n) = 30, 30|30. It works

Assume at n = k, 30|(k^5-k)

At n = k+1,
(k+1)^5-(k+1) mod 30
≡k^5+1-k+1 mod 30, because (k+1)^5 =∑(i, 0...5) C(5, i)k^(5-i)*
≡k^5-k mod 30
≡0 mod 30, because we assumed 30|(k^5-k)

*More explanation:
5K^4+10k^3+10k^2+5k mod 30
≡5k(k^3+1)+10k(k+1) mod 30
≡5k(k+1)(k^2-k+3) mod 30
≡5k(k+1)(k^2-k) mod 30
≡5k(k+1)(k-1)k mod 30
≡0 mod 30

End of proof.