Using euclids division lemma prove that for any positiv integer n,n3-n is divisible by 6
Answers
Answer:
Step-by-step explanation:
Any positive integer is of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5 for some positive integer n.
When n = 6m,
n3 - n = (6m)3 - 6m = 216 m3 - 6m
= 6m(36m2 - 1)
= 6q, where q = m(36m2 -1)
n3 - n is divisible by 6
When n = 6m + 1,
n3 - n = n(n2 - 1) = n (n - 1) (n + 1)
= (6m + 1) (6m) (6m + 2)
= 6m(6m + 1) (6m + 2)
= 6q, where q = m(6m + 1) (6m + 2)
n3 - n is divisible by 6
When n = 6m + 2,
n3 - n = n (n - 1) (n + 1)
= (6m + 2) (6m + 1) (6m + 3)
= (6m + 1) (36 m2 + 30m + 6)
= 6m (36 m2 + 30m + 6) + 1 (36m2 + 30m + 6)
= 6[m (36m2 + 30m + 6)] + 6 (6m2 + 5m + 1)
= 6p + 6q,
where p = m (36m2 + 30m + 6)
q = 6m2 + 5m + 1
n3 - n is divisible by 6
When n = 6m + 3
n3 - n = (6m + 3)3 - (6m + 3)
= (6m + 3) [(6m + 3)2 - 1]
= 6m [6m + 3)2 - 1] + 3 [(6m + 3)2 - 1]
= 6 [m [(6m + 3)2 - 1] + 3 [36m2 + 36m + 8]
= 6 [m [(6m + 3)2 - 1] + 6 [18m2 + 18m + 4]
= 6p + 3q,
where p = m[(6m + 3)2 - 1]
q = 18m2 + 18m + 4
n3 - n is divisible by 6
When n = 6m + 4
n3 - n = (6m + 4)3 - (6m + 4)
= (6m + 4) [(6m + 4)2 - 1]
= 6m [(6m + 4)2 - 1] + 4 [(6m + 4)2 - 1]
= 6m [(6m + 4)2 - 1] + 4 [36m2 + 48m + 16 - 1]
= 6m [(6m + 4)2 - 1] + 12 [12m2 + 16m + 5]
= 6p + 6q,
where p = m [(6m + 4)2 - 1]
q = 2 (12 m2 + 16m + 5)
n3 - n is divisible by 6
When n = 6m + 5
n3 - n = (6m + 5) [(6m + 5)2 - 1]
= 6m [(6m + 5)2 - 1] + 5 [(6m + 5)2 - 1]
= 6m [(6m + 5)2 - 1] + 5 [36m2 + 60m + 24]
= 6p + 30q
= 6 (p + 5q),
where p = m [(6m + 5)2 - 1]
q = 6m2 + 10m + 4
n3 - n is divisible by 6
Hence, n3 - n is divisible by 6, for any +ve integer n
hope it is clear
Answer:
n3 – n = n(n2 – 1) = n(n+1)(n – 1) = (n – 1)n(n+1) = product of three consecutive positive integers.
Now, we have to show that the product of three consecutive positive integers is divisible by 6.
We know that any positive integer n is of the form 3q, 3q + 1 or 3q + 2 for some positive integer q.
Now three consecutive positive integers are n, n + 1, n + 2.
Case I. If n = 3q.
n(n + 1) (n + 2) = 3q(3q + 1) (3q + 2)
But we know that the product of two consecutive integers is an even integer.
∴ (3q + 1) (3q + 2) is an even integer, say 2r.
⟹ n(n + 1) (n + 2) = 3q × 2r = 6qr, which is divisible by 6.
Case II. If n = 3n + 1.
∴ n(n + 1) (n + 2) = (3q + 1) (3q + 2) (3q + 3)
= (even number say 2r) (3) (q + 1)
= 6r (q + 1),
which is divisible by 6.
Case III. If n = 3q + 2.
∴ n(n + 1) (n + 2) = (3q + 2) (3q + 3) (3q + 4)
= multiple of 6 for every q
= 6r (say),
which is divisible by 6.
Hence, the product of three consecutive integers is divisible by 6.