Show that the cube of any postive in integer is of the form. 4m or (4m+1)or 4m+3 for some integer m.
(Hint n. 4q,4q+1,4q+2,4q+3
Answers
Let 'a' be any positive integer and b = 4.
Using Euclid Division Lemma,
a = bq + r [ 0 ≤ r < b ]
⇒ a = 3q + r [ 0 ≤ r < 4 ]
Now, possible value of r :
r = 0, r = 1, r = 2, r = 3
If we take, r = 0
⇒ a = 4q + 0
⇒ a = 4q
On cubing both sides,
⇒ a³ = (4q)³
⇒ a³ = 4 (16q³)
⇒ a³ = 9m [16q³ = m as integer]
If we take, r = 1
⇒ a = 4q + 1
On cubing both sides ;
⇒ a³ = (4q + 1)³
⇒ a³ = 64q³ + 1³ + 3 * 4q * 1 ( 4q + 1 )
⇒ a³ = 64q³ + 1 + 48q² + 12q
⇒ a³ = 4 ( 16q³ + 12q² + 3q ) + 1
⇒ a³ = 4m + 1 [ Take m as some integer ]
If we take r = 2,
⇒ a = 4q + 2
On cubing both sides ;
⇒ a³ = (4q + 2)³
⇒ a³ = 64q³ + 2³ + 3 * 4q * 2 ( 4q + 2 )
⇒ a³ = 64q³ + 8 + 96q² + 48q
⇒ a³ = 4 ( 16q³ + 2 + 24q² + 12q )
⇒ a³ = 4m [Take m as some integer]
If we take, r = 3
⇒ a = 4q + 3
On cubing both the sides;
⇒ a³ = (4q + 3)³
⇒ a³ = 64q³ + 27 + 3 * 4q * 3 (4q + 3)
⇒ a³ = 64q³ + 24 + 3 + 144q² + 108q
⇒ a³ = 4 (16q³ + 36q² + 27q + 6) + 3
⇒ a³ = 4m + 3 [Take m as some integer]
Hence, the cube of any positive integer is in the form of 4m, 4m+1 or 4m+3.