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!

We can expand n⁵ – n into factors:

n⁵ – n = n(n² + 1)(n + 1)(n - 1)  [1]

If n is even, n = 2k and we obtain: n⁵ – n = 2k(4k² + 1)(2k + 1)(2k - 1)  [2]. Therefore, n⁵ – n = M2 since [2] contains a factor of 2.

If n is odd, n = 2k + 1 and: n⁵ – n = (2k + 1)[(2k + 1).(2k+1)+1](2k + 2)(2k)   [3]. Therefore, n⁵ – n = M2 since [3] also contains a factor of 2.

The n can be M3, M3 + 1, or M3 - 1 and

If n = M3, n⁵ – n is necessarily M3 since its expansion [1] contains the factor n.

If n = M3 + 1, then n⁵ – n is necessarily M3 since its expansion [1] contains the factor (n-1) = M3.

If n = M3 - 1, then n⁵ – n is necessarily M3 since its expansion [1] contains the factor (n+1) = M3.

Therefore, for all n, n⁵ – n is divisible by 2 and 3, and therefore by 6.

n^5-n = n(n-1)(n+1)(n^2+1) 
n-1, n and n+1 are 3 consecutive numbers

for n = 0,1; product will always be zero (which is 6*0)
for all other n, one number amongst n, (n-1) and (n-2) has to be even number (multiple of 2) and one number has to be multiple of 3
Thus their product has to be divisible by 6

Factor (n⁵-n) into n(n-1)(n+1)(n²+1) then divide by 2x3.

Equation A:

 For some odd numbers, n/3 is an integer. If the remainder of n/3 is 1/3, then (n-1)/3 is an integer. If it is 2/3, then (n+1)/3 is an integer. The second term is an integer since (n²+1) is an even number.

Equation B:

For the first term, n/2 is an even integer. For the second term, If the remainder of n/3 is 1/3, then (n-1)/3 is an integer. If it is 2/3, then (n+1)/3 is an integer

Equation C:

For the first term, n/6 is an integer.

This covers the range of n >= 0 and shows that whatever n is, the result of (n⁵-n)/6 is an integer (i.e. it’s divisible by 6).

if we expand n power 5 - n then it will be n(n power 4 - 1)
which is (n square - 1)(n square + 1) => (n square + 1) (n + 1)(n - 1) => (n-1) *  n * (n +1)* (n square + 1)
except 0 and 1 (n-1) *  n * (n +1) will be three consecutive number so the product of three number alwasy divisble by 6 because divisibility rule of 2 and 3 Since at least one of the numbers in the product is even, it is divisible by 2, and one of the numbers is divisible by 3, the product must also be divisible by 3. Therefore, the product is divisible by 6
for 0 and 1 value is 0 so divisible by 0 also

(n^5 -1) = n(n-1)(n+1)(n^2 + 1)

n-1, n, n+1 are three consecutive integers. Therefore, one of them must be divisible by 3 and at least one must be divisible by 2. We can conclude that the product of these two integers must be divisible by 6.

N^5-n factors into n(n-1)(n+1)(N^2+1).  The first three terms will always have an even number and a multiple of 3,  So the entire function will have 6 as a factor.

For extra credit, prove that it is also divisible by 5 (and hence 30). 

Factors of 6 are 2 and 3, so if a number is divisible by both 2 and 3, it's divisible by 6.

First, tackle the divisible by 2 part.

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

For odd values of n, the square is also odd, square again and it's still odd.  Any odd number -1 is even, so the n^4-1 term is always divisible by 2 for odd n's.

For even values of n, the n outside  n(n^4-1) is always even so it is divisible by 2.

Now the divisible by 3 part.

All numbers can be represented by n = 3x, 3x+1, or 3x-1 for some value of x.

A) If it's a 3X number, then the n outside is divisble by 3.

B) When n = 3x+1 or 3x-1, you can show that n^4 - 1 is always divisible by 3:

  (3x+1) ^2 = 9x^2 + 6x + 1

Square that again for n^4 and you get 81x^4 + 108x^3 + 54x^2 + 12x + 1.

So n^4 - 1 = 3(27x^4 + 36x^3 + 18x^2 + 4x) which is always divisible by 3 due to the factor of 3 outside.

(3x-1)^4 power gives a very similar result, where all factors are multiples of 3 except the final +1 which is cancelled with the -1.

This shows that n(n^4 - 1) is always divisible by both 2 and 3, so divisible by 6.