# Investigating Divisibility

. Factorise the expression P(n) = nx - n for x ? {2, 3, 4, 5}. Determine if the expression is always divisible by the corresponding x. If divisible use mathematical induction to prove your results by showing whether P(k+1) - P(k) is always divisible by x. Using appropriate technology, explore more cases, summarize your results and make a conjecture for when nx - n is divisible by x.

.1 Factorise the expression P(n) = nx - n for x ? {2, 3, 4, 5}.

When n is real numbers;

x = 2 P(n) = n2 - n = n (n - 1)

x = 3 P(n) = n3 - n = n (n2 - 1) = n (n + 1) (n - 1)

x = 4 P(n) = n4 - n = n (n3 - 1) = n (n - 1) (n2 + n + 1)

x = 5 P(n) = n5 - n = n (n4 - 1) = n (n2 + 1) (n2 - 1)

= n (n2 + 1) (n - 1) (n + 1)

.2 Determine if the expression is always divisible by the corresponding x.

1.2.1 x = 2 P(n) = n2 - n = n (n - 1) divisible by 2?

When n = 1

n (n - 1) = 1 (1 - 1)

= 1 (0)

As 0÷2 = 0, P(n) is divisible by 2 when x = 2 and n = 1

Assume n = k is correct

k (k - 1) = k2 - k = 2M (where M is any natural number)

Then considering n = k + 1

(k + 1) (k) = k2 + k

= (k2 - k) +2k

= 2M + 2k

= 2(M + k)

Therefore, P(n) = n2 - n is divisible by 2 when x = 2 and n = any natural numbers.

1.2.2 x=3 P(n) = n (n + 1) (n - 1) divisible by 3?

When n = 1

n (n + 1) (n - 1) = 1 (2) (0)

= 0

As 0÷3 = 0, P(n) is divisible by 3 when x = 3 and n = 1

Assume n = k is correct

k (k + 1) (k - 1) = k (k2 - 1)

= k3 - k

= 3M (where M s any natural number)

Then considering n = k + 1

(k + 1) {(k + 1) + 1} {(k + 1) - 1} = (k + 1) (k + 2) (k)

As there is a multiplication of three consecutive integers, one of these three will always be a multiple of 3. Hence when x =3, P(n) = n (n + 1) (n - 1) is divisible by 3.

1.2.3 x = 4 P(n) = n (n - 1) (n2 + n + 1)

When n = 1

n (n - 1) (n2 + n + 1) = 1(0)(1 + 1 + 1)

= 0

As 0÷4 = 0, P(n) is divisible by 4 when n = 1.

Assume n = k is correct

k (k - 1) (k2 + k + 1) = 4M (where M is any natural number)

Considering n = k+1

(k + 1) {(k + 1) - 1} {(k + 1)2 + (k + 1) + 1} = (k + 1) (k) (k2 + 3k + 3)

As there is no 4 as a coefficient, which means it is not a multiple of four consecutive numbers, therefore when x = 4, P(n) = n (n - 1) (n2 + n + 1) is not divisible by 4.

1.2.4 x = 5 P(n) = n (n2 + 1) (n - 1) (n + 1)

When n = 1

n (n2 + 1) (n - 1) (n + 1) = 1 (2) (0) (2)

= 0

As 0÷5 is 0, P(n) is divisible by 5 when n = 1.

Assume n = k is correct

n (n2 + 1) (n - 1) (n + 1) = 5M (where M is any natural number)

Considering n = k + 1

(k + 1) {(k + 1)2 + 1} {(k + 1) - 1} {(k + 1) + 1} = (k + 1) (k2 + 2k + 2) (k) (k + 2)

= (k3 + 3k2 + 2k) (k2 + 2k + 2)

= k5 + 5k4 + 10k3 + 10k2 + 4k

As there is coefficient that is multiples of 5, when x = 5, P(n) = n (n2 + 1) (n - 1) (n + 1) is divisible by 5 when k is any natural number.

.3 If divisible use mathematical induction to prove your results by showing whether P(k+1) - P(k) is always divisible by x.

As P(n) = nx - n P(k+1) - P(k) = (k + 1)x - (k + 1) - kx + k

1.3.1 x = 2

P(k+1) - P(k) = (k + 1)x - (k + 1) - kx + k

= (k + 1)2 - (k + 1) - k2 + k

= k2 + 2k + 1 - k - 1 - k2 + k

= 2k

Therefore, as P(n) = nx - n and x = 2, P(k+1) - P(k) is divisible by x.

1.3.2 x = 3

P(k+1) - P(k) = (k + 1)x - (k + 1) - kx + k

= (k + 1)3 - (k + 1) - k3 + k

= k3 + 3k2 + 3k + 1 - k - 1 - k3 + k

= 3k2 + 3k

= 3(k2 + k)

Therefore, as P(n) = nx - n and x = 3, P(k+1) - P(k) is divisible by x.

1.3.3 x = 4

P(k+1) - P(k) = (k + 1)x - (k + 1) - kx + k

= (k + 1)4 - (k + 1) - k4 + k

= k4 + 4k3 + 6k2 + 4k + 1 - k - 1 - k4 + k

= 4k3 + 6k2 + 4k

For P(n) = nx - n and x ...