Prompt Title: proving discrete math

To show that n^2 - n is divisible by 2, we can factor out n from the expression as follows:

n^2 - n = n(n-1)

Either n or (n-1) is even for all integers n, so the product of n and (n-1) is even, and therefore n^2 - n is divisible by 2.

To show that n^3 - n is divisible by 6, we can factor out n as before:

n^3 - n = n(n^2 - 1)

Now, n^2 - 1 can be factored further as (n+1)(n-1). One of these factors is even and the other is divisible by 3 for all integers n, so their product is divisible by 6. Therefore, n^3 - n is divisible by 6.

Finally, to show that n^5 - n is divisible by 30, we can again factor out n:

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

Similar to the previous case, n^4 - 1 can be factored as (n^2 + 1)(n^2 - 1), and one of these factors is divisible by 3 and the other by 5 for all integers n, so their product is divisible by 15. Also, one of these factors is even and the other is odd, so their product is divisible by 2. Therefore, n^5 - n is divisible by 30, which is the least common multiple of 2 and 15.