# prove that square of any odd integer is of the form 8m+1,for some integer m

prove that square of any odd integer is of the form 8m+1,for some integer m

- Asked in: CBSE Class 10
- Asked By prashant
- Asked on September 25, 2013

prove that square of any odd integer is of the form 8m+1,for some integer m

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## Answers

let the odd integer be in the form of 4m+1,4m+3………………………………

on squaring

(4m+1)^2

=16m^2+1+8m

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

=8m+1(where m=2m^2+m)

done

Its so simple

Lets start

Solution: let ‘a’ is an any possitive integer when ‘a’ is divided by 8 it gives quotent ‘m’ and remainder ‘r’.

Now,

a=8m

Or, a=2(4m) it is clearly even number because 2 is its multiple and remainder are leaves after 4m.

Now,

a=8m 1

It is clearly an odd number because it we put out 2 like a=2(4m) 1 then here 1 is rest as remainder so the term 8m 1 is and odd integer.

Let ‘a’ be given positive odd integer.

If we divide ‘a’ by 4 then by Euclid’s DA

a = 4q + r where 0_< r<4

So r may be 0, 1, 2, 3

putting 1 in place of r

a=4q+1

a^2= (4q+1)^2

a^2=16q^2+8q+1

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

a^2= 8m+1 where 'm' is equal to q(q+1)

hence square of a positive odd integer is of form 8m +1 for some integer 'm'………………