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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Step-by-step explanation:
let ' a' be any positive integer and b = 3.
we know, a = bq + r , 0 < r< b.
now, a = 3q + r , 0<r < 3.
the possibilities of remainder = 0,1 or 2
Case I - a = 3q
a² = 9q² .
= 3 x ( 3q²)
= 3m (where m = 3q²)
Case II - a = 3q +1
a² = ( 3q +1 )²
= 9q² + 6q +1
= 3 (3q² +2q ) + 1
= 3m +1 (where m = 3q² + 2q )
Case III - a = 3q + 2
a² = (3q +2 )²
= 9q² + 12q + 4
= 9q² +12q + 3 + 1
= 3 (3q² + 4q + 1 ) + 1
= 3m + 1 ( where m = 3q² + 4q + 1)
From all the above cases it is clear that square of any positive integer ( as in this case a² ) is either of the form 3m or 3m +1.
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