Show-that-the-square-of-every-odd-integer-is-of-the-form-8m-1-

Question Number 45439 by Tawa1 last updated on 12/Oct/18

Show that the square of every odd integer is of the form 8 m + 1

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18

{(2 k + 1)}^{2} = 4 k^{2} + 4 k + 1

= 4 k (k + 1) + 1

n o w p u t k = 1 8 + 1 i n t h e f o r m 8 \times 1 + 1

p u t k = 2 24 + 1 i n t h e f o r m 8 \times 3 + 1

n o w a s s u m e f o r k = r t h e s t a t e m e n t i s t r u e

t h a t i s 4 r (r + 1) + 1 i n t h e f o r m 8 m + 1

t o s h o w i t [i s t r u e w h e n k = r + 1

p u t k = r + 1

4 (r + 1) (r + 2) + 1

= 4 (r + 1) (r + 1 + 1) + 1

= 4 {r (r + 1) + r + r + 1 + 1} + 1

4 r (r + 1) + 4 (2 r) + 8 + 1

4 r (r + 1) + 8 (r + 1) + 1

f o r m o f 8 m + 1 + 8 (r + 1) + 1

Commented by Tawa1 last updated on 13/Oct/18

God bless you sir

Answered by MJS last updated on 13/Oct/18

odd integer o = 2 n + 1

o^{2} = {(2 n + 1)}^{2}

n = 2 k \lor n = 2 k + 1

n = 2 k

o^{2} = {(4 k + 1)}^{2} = 16 k^{2} + 8 k + 1 =

= 8 (2 k^{2} + k) + 1 = 8 m + 1

k = \frac{n}{2} \Rightarrow m = k (2 k + 1) = \frac{n (n + 1)}{2}

n = 2 k + 1

o^{2} = {(4 k + 3)}^{2} = 16 k^{2} + 24 k + 9 =

= 8 (2 k^{2} + 3 k + 1) + 1 = 8 m + 1

k = \frac{n - 1}{2} \Rightarrow m = (k + 1) (2 k + 1) = \frac{n (n + 1)}{2}

Commented by Tawa1 last updated on 13/Oct/18

God bless you sir

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