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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.
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