Question

show that cube of any positive integer is of form 4m 4m+1 4m+3 for some integer m

Answer 1

Step-by-step explanation:

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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Answer 2

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.Read more on Sarthaks.com - https://www.sarthaks.com/12812/show-that-cube-of-any-positive-integer-is-of-the-form-4m-4m-1-or-4m-3-for-some-integer-m