Question

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

Answer 1

Let a be an arbitrary positive integer.

Then, by division algorithm, 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

`\implies` a³ = (4q + r)³

= 64q³ + r³ + 12qr² + 48q²r   .......[∵ (a + b)³ = a³ + b³ + 3ab² + 3a²b)

`\implies` a³ = (64q³ + 48q²r + 12qr²) + r³  .......(i)

Where, 0 ≤ r < 4

Case I: When r = 0,

Then putting r = 0 in equation (i), we get

a³ = 64q³ = 4(16q³)

`\implies` a³ = 4m, where, m = 16q³ is an integer.

Case II: When r = 1,

Then putting r = 1 in equation (i), we get

a³ = 64q³ + 48q² + 12q + 1

= 4(16q³ + 12q² + 3q) + 1

= 4m + 1

Where, m = (16q³ + 12q² + 3q) is an integer.

Case III: When r = 2,

Then putting r = 2 in equation (i), we get

a³ = 64q³ + 96q² + 48q + 8

= 4(16q³ + 24q² + 12q + 2)

= 4m

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

Case IV: When r = 3,

Then putting r = 3 in equation (i), we get

a³ = 64q³ + 144q² + 108q + 27

= 64q³ + 144q² + 108q + 24 + 3

= 4(16q³ + 36q² + 27q + 6) + 3

= 4m + 3

Where, m = (16q³ + 36q² + 27q + 6) is an integer.

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