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Question

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, but not of the form 3m+2.

Solution
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Let a be the positive integer and b=3.

We know a=bq+r, 0r<b

Now, a=3q+r, 0r<3

The possibilities of remainder is 0,1, or 2.

Case 1 : When a=3q

a2=(3q)2=9q2=3q×3q=3m where m=3q2

Case 2 : When a=3q+1

a2=(3q+1)2=(3q)2+(2×3q×1)+(1)2=3q(3q+2)+1=3m+1 where m=q(3q+2)

Case 3: When a=3q+2

a2=(3q+2)2=(3q)2+(2×3q×2)+(2)2=9q2+12q+4=9q2+12q+3+1=3(3q2+4q+1)+1=3m+1
where m=3q2+4q+1

Hence, from all the above cases, it is clear that square of any positive integer is of the form 3m or 3m+1.

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