KCC's Quiz AQQ300 about Divisibility by 6

To show that   ( n^5 – n )   is divisible by 6 for any integer n >= 0,
we can use a mix of factoring and logic.
For a number to be divisible by 6, it must be divisible by both 2 and 3.

1. Factor the expression

First, we break the expression down into smaller parts:
( n^5 – n ) = n (n^4 - 1)
= n (n^2 - 1) (n^2 + 1)
= n (n - 1) (n + 1) (n^2 + 1)

If we rearrange the first three terms, we get:
(n - 1) n (n + 1) (n^2 + 1)

2. Check for divisibility by 2 and 3

Now we look at the product of the three consecutive integers:
(n - 1) n (n + 1)

Divisibility by 2: In any two consecutive integers like    (n - 1) n
one must be even. Therefore, the product is always divisible by 2.

Divisibility by 3: In any three consecutive integers, exactly one must be a multiple of 3. Therefore, the product   (n-1) n (n+1)   is always divisible by 3.

3. Conclusion

Since the expression is always divisible by both 2 and 3,
it must be divisible by   2 x 3 = 6.

Brillant steve12 ! You found the solution and with a very clear explanation, co,gratulations!

Brillant steve12 ! You found the solution and with a very clear explanation, co,gratulations!