Question

show that the cube of any positive integer is of the form 4m+1 or 4m+3 where m is whole number​

Answer 1

Reffer the attachment friend.....☝︎☝︎☝︎

Answer 2

Answer:

Let a be an arbitrary positive integer. Then, by Euclid’s 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

a = 4q + r, where 0 ≤ r < 4

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

[∵(a+b)3 = a3 + b3 + 3ab2 + 3a2b]

⇒ a3 = (64q2 + 48q2r + 12qr2) + r3

where, 0 ≤ r < 4

Case I When r = 0,

Putting r = 0 in Eq.(i), we get

a3 = 64q3 = 4(16q3)

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

Case II When r = 1, then putting r = 1 in Eq.(i), we get

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

= 4(16q3 + 12q2 + 3q) + 1

= 4 m + 1

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

Case III When r = 2, then putting r = 2 in Eq.(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.