Question

show that cube of any positive integer is of the form 4m, 4m+1 or 4m+3,for some integer​

Answer 1

Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer.

Let a be an arbitrary positive integer. Then corresponding to the positive integers a and 4, there exist non-negative integers q and r such that

a = 4q + r, where 0 ≤ r < 4

Cubing both the sides using (a + b)3 = a3 + b3 + 3ab2 + 3a2b

⇒ a3 = (4q + r)3 = 64q3 + r3 + 12qr2 + 48q2r

⇒ a3 = (64q3 + 48q2r + 12qr2) + r3

where 0 ≤ r < 4

Case I When r = 0.

a3 = 64q3 = 4(16q3)

⇒ a3 = 4m where m = 16q3 is an integer.

Case II When r = 1 we get

a3 = 64q3 + 48q2 + 12q + 1

⇒ a3 = 4 (16q3 + 12q2 + 3q) + 1

⇒ a3 = 4m + 1

where m = (16q2 + 12q2 + 3q) is an integer.

Case III When r = 2 we get

a3 = (64q3 + 96q2 + 48q) + 8

⇒ a3 = 4 (16q3 + 24q2 + 12q + 2)

⇒ a3 = 4m

where, m = (16q3 + 24q2 + 12q + 2) is an integer.

Case IV When r = 3 we get

a3 = (64q3 + 144q2 + 108q) + 27

⇒ a3 = (64q3 + 144q2 + 108q) + 24 + 3

⇒ a3 = 4 (16q3 + 36q2 + 27q + 8) + 3

⇒ a3 = 4m + 3

where m = (16q3 + 36q2 + 27q + 8) is an integer.

Hence, of any positive integer is of