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

By 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 the form 4m, 4m + 1 or 4m + 3, for some integer m.