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

3

Answers


  1. 0 Votes

    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

  2. 0 Votes

    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.

  3. 0 Votes

    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'………………

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