Use division algorithm to show square of any positive integer is of form 3p Or 3p+1
Answers
Let A be any square of positive integer and b=3 by using Euclid division lemma A=B×Q+R
the possible remainders are :0,1,2
case-1:
r=0
a=3q+r
a=3q+0
squaring on both sides
a^2=(3q+0)^2
a^2=(3q)^2
a^2=9q^2
a^2=3(3q^2). (3would be common)
a^2=3(p). ( where p= 3q^2)
case-2:
r=1
a=3q+1
squaring on both sides
a^2=(3q+1)^2
a^2=9q^2+1+2(3q×1). (a+b)^2=a^2+b^2+2(ab)
a^2=9q^2+1+6q
a^2=9q^2+6q+1
a^2=3(3q^2+2)+1. ( 3would be common)
a^2=3p+1. (where p=3q^2+2)
case-3:
r=2
a=(3q+2)
squaring on both sides
a^2=(3q+2)^2
a^2=9q^2+2(3q×2)+4. (a+b)^2=a^2+b^2+2(ab)
a^2=9q^2+12q+3+1
a^2=3(3q^2+4q+1)+1. (3would be common)
a^2=3(p)+1. (where p=3q^2+2)
so hence proved
Answer:
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.
Hence proved.