Question

show that square of any odd integer is of the form 4q+1 for some integer q.​

Answer 1

Answer:

let, the odd integer is q

so,

${q}^{2} \: is \: an \: odd \: integer$

now,

$q \: is \: odd \: \\ \: 4q \: is \: even$

as 4q is even, (4q+1) must be odd

here,

$both \: {q}^{2} \: and \: (4q + 1) \: are \: odd$

therefore the statement is true

Answer 2

$\huge\underline\underline\mathbb\blue{ANSWER}$

By Euclid's Division Lemma,

$a = bq + r \: \: \: where \: 0 \leqslant r < b$

Here,

$a = 4q + 1 \: \: \: and \: \: \: b = 4$

a = {1, 3, 5, 7...}

So, the Odd integers are,

$\mathbb{{1}^{2} = 1 = 4(0) + 1 }$

$\mathbb{{3}^{2} = 9 = 4(2) + 1 }$

$\mathbb{{5}^{2} = 25 = 4(6) + 1 }$

$\mathbb{{7}^{2} = 49 = 4(12) + 1 }$

$\mathbb\therefore$The square of any odd integer is of the form 4q+1.