Use Euclid's division algorithm to show that the cube of any positive
integer is of the form 4m or 4m +1 or 4m +3.
Answers
Step-by-step explanation:
Given:-
Cube of a positive integer
To find:-
Use Euclid's division algorithm to show that the cube of any positive integer is of the form 4m or 4m +1 or 4m +3.
Solution:-
We know that
For two positive integers a and b, then there exist unique integers q and r which satisfying a = bq + r where 0 ≤ r < b.
Let a = 4q+r, 0≤r<4
The possible values of r = 0,1,2,3
Case-1:-
If r = 0 then a= 4q+0
=> a = 4q
On cubing both sides then
=> a^3 = (4q)^3
=> a^3 = 64q^3
=> a^3 = 4(16q^3)
=> a^3 = 4m ------------(1)
(Where m = 16q^3)
Case -2:-
If r = 1 then a = 4q+1
On cubing both sides then
=> a^3 = (4q+1)^3
We know that
(a+b)^3 = a^3+3a^2b+3ab^2+b^3
=> a^3 = (4q)^3+3(4q)^2(1)+3(4q)(1)^2+(1)^3
=> a^3 = 64q^3+3(16q^2)+12q+1
=> a^3 = 64q^3+48q^2+12q+1
=>a^3= 4(16q^3+12q^2+3q)+1
=>a^3=4m+1 ---------------(2)
Where, m = 16q^3+12q^2+3q
Case -3:-
If r = 2 then a = 4q+2
on cubing both sides then
=> a^3 = (4q+2)^3
We know that
(a+b)^3 = a^3+3a^2b+3ab^2+b^3
=> a^3 = (4q)^3+3(4q)^2(2)+3(4q)(2)^2+2^3
=>a^3 = 64q^3+3(16q^2)(2)+3(4q)(4)+8
=>a^3 = 64q^3+96q^2+48q+8
=>a^3 = 4(16q^3+24q^2+12q+2)
=>a^3 = 4m ------------------(3)
Where m = 16q^3+24q^2+12q+2
Case -4:-
If r = 3 then a = 4q+3
On cubing both sides then
=> a^3 = (4q+3)^3
We know that
(a+b)^3 = a^3+3a^2b+3ab^2+b^3
=> a^3 = (4q)^3+3(4q)^2(3)+3(4q)(3)^2+(3)^3
=>a^3 = 64q^3+3(16q^2)(3)+3(4q)(9)+27
=>a^3 = 64q^3+144q^2+108q+27
=>a^3 = 64q^3+144q^2+108q+24+3
=>a^3 = 4(16q^3+32q^2+27q+6)+3
=>a^3 = 4m+3----------------(4)
Where, m = 16q^3+32q^2+27q+6
From above all equations
We conclude that
"The cube of any positive integer is in the form of either 4m or 4m+1 or 4m+3"
Hence, Proved.
Used formulae:-
Euclid's Division Lemma:-
For two positive integers a and b, then there exist unique integers q and r which satisfying
a = bq + r where 0 ≤ r < b.
- (a+b)^3 = a^3+3a^2b+3ab^2+b^3
This is a answer........... .......
