Question

show that the square of an odd positive integer is of the from 8m+1for some whole no. m

Answer 1

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'…