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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.

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

Answers (1)

By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 4
r is remainder when we divide A by 4, therefore:
 0\leq\ r  <4  r = 0, 1, 2, 3
A = 4q + r                   …(1)

Case 1:
A = 4q            
A3 = (4q)3 = 64q3 = 4(16q3)
A3 = 4m          
(m = 16 q3)

Case 2:
A = 4q + 1
  (4q + 1)3 = (4q)3 + (1)3 + 3(4q)2 (1) + 3(4q) (1)2
(Using (a + b)3 = a3+ b3 + 3a2b + 3ab2)
= 64q3 + 1 + 48q2 + 12q
= 4(16q3 + 12q2 + 3q) + 1
A3 = 4m + 1   
(m = 16q3 + 12q2 + 3q)

Case 3:
A = 4q + 2
  = (4q + 2)3 = (4q)3 + (2)3 + 3(4q)2 (2) + 3(4q) (2)2
(using (a + b)3 = a3 + b3 + 3a2b + 3ab2)
= 64q3 + 8 + 96q2 + 48q
= 4(16q3 + 2 + 24q2+ 12q)
= 4m   
(m = 16q3 + 2 + 24q2 + 12 q)

Case 4:
A = 4q + 3
A3 = (4q + 3)2 = (4q)3 + (3)3 + 3(4q)2 (3) + 3(4q) (3)2
(Using (a + b)3 = a3 + b3 + 3a2b + 3ab2)
= 64q3 + 27 + 144q2 + 108q
= 64q3 + 24 + 144 q2 + 108q + 3
= 4(16q3 + 6 + 36q2 + 27q) + 3
  = 4m + 3                     
(m = 16q3 + 6 + 36q2 + 27q)
Hence any positive integer’s cube can be written in the form of
4m, 4m + 1, 4m + 3.

Posted by

infoexpert27

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