Question

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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.



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Answer 1

let us consider a positive integer a

divided the positive integer a by 3, and let r be the remainder and B be the quotient and such that.

a=3b+r.........(1)

where =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 x 3b²

a²=3m

• where m=3b²
• case 2: Let R=1

• equation (1) become

a=3b+1

• squaring on both the side we get

• a²=(3b+1)²
• a²=(3b) ²+1+2x(3b)x1
• a²=9b²+6b+1
• a²=3(3b²+2b) +1

a²=3m+1

• where m=3b²+2b
• case3:let r=2
• equation (1)becomes

a=3b+2

• squaring on both the sides we get
• a²(3b+2)²
• a²=9b²+4(12x3bx2
• a²9b²+(12b+3+1)
• a²3(3b²+4b+1)+1

a²3m+1

where m=3b²+4b+1

: square of any positive integer is of the form 3M or 3M + 1

: hence proved.

Answer 2

It is the correct answer.

Explanation:

Hope this attachment helps you.