Question

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

Answer 1

Hey there!

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.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.


Hope it help

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.