Use uclied's devision lemma to show that the square of any positive integer is either of the form 3m or 3m +1 for some integer m
Answers
Step-by-step explanation:
Given :-
A square of any positive integer
To find :-
Use Euclids Division Lemma to show that the square of any positive integer is either of the form 3m or 3m +1 for some integer m ?
Solution :-
Euclid's Division Lemma:-
For any two integers a and b there exist two integers q and r satisfying a=bq+r, 0≤r<b.
Let Consider a = 3q+r, 0≤r<3------(1)
The possible values of r = 0,1,2.
If r = 0 then (1) becomes
=> a = 3q+0
=> a = 3q
On squaring both sides
=> a^2 = (3q)^2
=> a^2 = 9q^2
=>a^2 = 3(3q^2)
=>a^2 = 3m -----------(2)
Where, m = 3q^2
and
If r = 1 then (1) becomes
=> a = 3q+1
on squaring both sides
=>a^2 = (3q+1)^2
=> a^2 = (3q)^2+2(3q)(1)+(1)^2
Since (a+b)^2 = a^2+2ab+b^2
=>a^2 = 9q^2+6q+1
=>a^2 = 3(3q^2+2q)+1
=> a^2 = 3m+1------------(3)
Where m = 3q^2+2q
If r = 2 then (1) becomes
=> a = 3q+2
On squaring both sides
=> a^2 = (3q+2)^2
=> a^2 = (3q)^2+2(3q)(2)+(2)^2
Since (a+b)^2 = a^2+2ab+b^2
=> a^2 =9q^2+12q+4
=>a^2 = 9q^2+12q+3+1
=>a^2=3(3q^2+4q+1)+1
=> a^2 = 3m+1 --------------(4)
Where m = 3q^2+4q+1
From (2),(3)&(4)
We conclude that
The square of any positive integer is either of the form 3m or 3m +1 for some integer m
Hence , Proved.
Used formulae:-
- (a+b)^2 = a^2+2ab+b^2
Euclid's Division Lemma:-
For any two integers a and b there exist two integers q and r satisfying a=bq+r, 0≤r<b.