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. Please explain in detail.

Answer 1

As per Euclid's Division Lemma
If a & b are 2 positive integers, then
$a = bq + r$
$where \: 0 \leqslant r < b$
Let positive integer be a
And b = 3
$hence \: a = 3q + r$
$where(0 \leqslant r < 3)$
r is an integer greater than or equal to 0 and less than 3
hence, r can be either 0, 1 or 2.

CASE 1:-
$if \: r = 0 \\ our \: equation \: becomes \\ a = 3q + r \\ a = 3q + 0 \\ a = 3q \\ squaring \: both \: the \: sides \\ {a}^{2} = {(3q)}^{2} \\ {a}^{2} = 9 {q}^{2} \\ {a}^{2} = 3(3 {q}^{2} ) \\ {a}^{2} = 3m \\ where \: m = 3 {q}^{2}$

CASE 2:-
$if \: r = 1 \\ our \: equation \: becomes \\ a = 3q + r \\ a = 3q + 1 \\ squaring \: both \: the \: sides \\ {a}^{2} = {(3q + 1)}^{2} \\ {a}^{2} = {(3q)}^{2} + {1}^{2} + 2(3q) \\ {a}^{2} = 9 {q}^{2} + 6q + 1 \\ {a}^{2} = 3 ({3q}^{2} + 2q) + 1 \\ {a }^{2} = 3m + 1 \\ where \: m \: = 3 {q}^{2} + 2q$


CASE 3:-
$if \: r \: = 2 \\ our \: equation \: becomes \\ a = 3q + r \\ a = 3q + 2 \\ squaring \: both \: the \: sides \\ {a}^{2} = {(3q + 2)}^{2} \\ {a}^{2} = {(3q)}^{2} + {2}^{2} + 2(2)(3q) \\ {a}^{2} = 9 {q}^{2} + 12q + 4 \\ {a}^{2} = 9 {q}^{2} + 12q + 3 + 1 \\ {a}^{2} = 3(3 {q}^{2} + 4q + 1) + 1 \\ {a}^{2} = 3m + 1 \\ where \: m = 3 {q}^{2} + 4q + 1$
Hence, square of any positive number can be expressed of the form 3m or 3m + 1

HENCE PROVED
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Answer 2

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.

Hence, it is solved .