Question

Show that square of any any positive odd integer is of the form 8m+1,for some whole number m

Answer 1

Hey!

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According to Euclid's Lemma,
We have two give positive integers and there always exist two unique integers 'q' and 'r' which satisfies the -:

a = bq + r

Where, 0 ≤ r < b.

Let's take,

b = 8
a = 8q + 1r

r will have value till 7 , (r < b).
r = 1, 2 , .... 7


CASE - 1 :-

r = 0

a = bq + r

a = 8q + 0

a = 8q

Squaring both sides =>

a^2 = (8q)^2

a^2 = 64 q^2

a^2 = 8(8q^2)

= 8 m

[m = 8q^2 ]

CASE - 2 -:


r = 1

a = bq + r

a = 8q + 1

Squaring both sides =>

a^2 = (8q + 1)^2

a^2 = [ (8q)^2 + 2 × 8q × 1 + 1^2)]

a^2 = 64 q^2 + 16q + 1

= 8 (8q^2 +2q) + 1

= 8m + 1

[ m = 8q^2 + 2q]


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Hope it helps...!!!

Answer 2

hey mate

here is the answer

let a be any positive odd integer then b= 8 ..





answer is in pic

hope it may help you

regards utsav
from brainly

by brainly

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