Question

Prove that the square of an odd number is of the form 8k+1

Answer 1

Question:

Prove that the square of an odd number is of the form 8k+1

Solution:

Let b an odd integer.

Then b is of the form 4k+1 or (-4k+1) , where k ∈ Z

${b}^{2} = (±4k + 1)^{2} \\ \\ \: \: \: \: \: = 16 {k}^{2} + 8k + 1 \\ \\ \: \: \: \: = 8(2 {k}^{2} + k) + 1 \\ \\ \: \: \: \: = 8q + 1 \:$

Where

$q = 2 {k}^{2} + k$

and q ∈ Z

Hence the square of an odd integer is of the form 8q+1 where q ∈ Z.

Answer 2

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✨$\underline \mathbb{SOLUTION}$

✨$8k + 1$

✨$\underline \mathbb{LET}$

✨B be an odd integer.

✨Thus,b is of the form 4k + 1 or -4k+1 where k belongs to Z.

✨${b}^{2} = ( 4k + 1) {}^{2}$

✨$= > 16 {k}^{2} + 8k + 1$

✨$= > 8(2 {k}^{2} + k) + 1$

✨$= > 8q + 1$

✨$\underline \mathbb{WHERE}$

✨$q = 2 {k}^{2} + k$

✨$and \: q \: belongs \: to \: z$

✨So, the square of an odd number is of the form 8k+1.

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