Use Euclid's division 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 positive integer
To find :-
Use Euclid's 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 :-
We know that
Euclid's Division Lemma:
For any two positive integers a and b there exists two positive integers q and r satisfying a = bq+r where ,0≤r<b.
Let a be any positive integer
and b = 3 then
On writing a = bq+r
=> a = 3q+r, 0≤r<3 ---------(1)
The possible values of r = 0,1,2.
Case -1:-
If r = 0 then (1) becomes
=> a = 3q+0
=> a = 3q
On squaring both sides then
=> a² = (3q)²
=> a² = 9q²
=> a² = 3(3q²)
=> a² = 3m -----------(2)
Where m = 3q² is an integer
Case -2:-
If r = 1 then (1) becomes
=> a = 3q+1
On squaring both sides then
=> a² = (3q+1)²
=> a² = (3q)²+2(3q)(1)+1²
Since (a+b)² = a²+2ab+b²
=> a² = 9q²+6q+1
=> a² = 3(3q²+2q)+1
=> a² = 3m +1 -----------------(3)
Where m = 3q²+2q ,is an integer
Case -3:-
If r = 2 then (1) becomes
=> a = 3q+2
On squaring both sides then
=> a² = (3q+2)²
=> a² = (3q)²+2(3q)(2)+2²
Since (a+b)² = a²+2ab+b²
=> a² = 9q²+12q+4
=> a² = 9q²+12q+3+1
=> a² = 3(3q²+4q+1)+1
=> a² = 3m+1 ----------------(4)
Where ,m = 3q²+4q+1 is an integer.
From (2),(3),&(4)
The square of any positive integer is in the form of 3m or 3m+1 .
Hence , Proved.
Answer:-
The square of any positive integer is either of the form 3m or 3m+1 for some integer m.
Used formulae:-
Euclid's Division Lemma:
For any two positive integers a and b there exists two positive integers q and r satisfying a = bq+r where ,0≤r<b.
- (a+b)² = a²+2ab+b²
Let us consider a positive integer a⤵️
Divide the positive integer a by 3, and let r be the reminder and b be the quotient such that
a = 3b + r……………………………(1)
where r = 0,1,2,3…..
Case 1: Consider r = 0
Equation (1) becomes
a = 3b
On squaring both the side
a² = (3b)²
a² = 9b²
a² = 3 × 3b²
a² = 3m
Where m = 3b²
Case 2: Let r = 1
Equation (1) becomes
a = 3b + 1
Squaring on both the side we get
a² = (3b + 1)²
a² = (3b)² + 1 + 2 × (3b) × 1
a² = 9b² + 6b + 1
a² = 3(3b² + 2b) + 1
a² = 3m + 1
Where m = 3b² + 2b
Case 3: Let r = 2
Equation (1) becomes
a = 3b + 2
Squaring on both the sides we get
a² = (3b + 2)²
a² = 9b² + 4 + (2 × 3b × 2)
a² = 9b² + 12b + 3 + 1
a² = 3(3b² + 4b + 1) + 1
a² = 3m + 1
where m = 3b² + 4b + 1
∴ Square of any positive integer is of the form 3m or 3m+1.