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

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

by (24.0k points)
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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.

by (10 points)
How to assume value of “b”? Here it is assumed as a=4q+0,1,2,3.
But in some similar problems value of b is taken as 3. So what is the criteria to assume value of b.