Question

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

Answer 1

Solution:
Let a be the positive integer and b = 4.
Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.
So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.
(4q)³ = 64q³ = 4(16q³)
= 4m, where m is some integer.
(4q + 1)³ = 64q³ + 48q² + 12q + 1
= 4(16q³ + 12q² + 3) + 1
= 4m + 1, where m is some integer.
(4q + 2)³ = 64q³ + 96q² + 48q + 8
= 4(16q³ + 24q² + 12q + 2)
= 4m, where m is some integer.
(4q + 3)³ = 64q³ + 144q² + 108q + 27
= 4(16q³ + 36q² + 27q + 6) + 3
= 4m + 3, where m is some integer.
Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.