show that the cube of any positive integer is of the form 4m, 4m+1, 4m+3 for some integer m
Answers
Answer:
Step-by-step explanation:
Let a,b be positive integers and b=4
then using euclid lemma
a=b*q+r,where q and r are any positive integers and 0<=r<b
∴r=0,1,3,2
so a=4q;a=4q+1;a=4q+2;a=4q+3
cubing
a^3=64q^3;
a^3=64q^3+1+12q(4q+1);
a^3=64q^3+8+24q(4q+2);
a^3=64q^3+27+36(4q+3)
a^3=4(16q^3);
a^3=4(16q^3+3q(4q+1))+1;
a^3=4(16q^3+2+6(4q+2));
a^3=4(16q^3+6+9(4q+3))+3;
Let 16q^3,16q^3+3q(4q+1),16q^3+2+6(4q+2),16q^3+6+9(4q+3) be m
∴a^3=4m;4m+1;4m;4m+3
=4m;4m+1;4m+3
Step-by-step explanation:
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
CASE 1:
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]
CASE 2:
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 ]
CASE 3:
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]
CASE 4 :
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.
__________________
Identity used ;
∵ ( a + b )³ = a³ + b³ + 3ab ( a + b ) .
Hence, it is solved.