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.


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