Question

Answer this question Fastly I will be mark you as brainlist.

Answer 1

let the positive integer be 'a'.
to prove,
a²=3m or 3m+1
let a be divided by 3
=>b=3
$0 \leqslant r < 3$
=>r=0,1,2
a=bq+r (by remainder theorem)
a=3q+r
a=3q,3q+1,3q+2
a²=(3q)²=9q²=3(3q²)=3m [m=3q²]
a²=(3q+1)²=9q²+6q+1=3(3q²+2q)+1=3m+1
a²=(3q+2)²=9q²+12q+4=9q²+12q+3+1=3(3q²+4q+1)+1=3m+1

therefore, square of any positive integer is of the form 3m or 3m+1.

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Answer 2

Hola mate..♥

Answer:-

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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