Question

show that the square of any +ve odd integer is of the form 8m+1​

Answer 1

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We know that any positive odd integer is of the form 4q+1 or 4q+3.

let a be any positive integer,then:

case 1: a=4q+1

a^2= (4q+1)^2 = 16q^2+8q+1

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

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

case 2: a=4q+3

a^2=(4q+3)^2 =16q^2+24q+9

= 8(2q^2+3q+1)+1

=8m+1 ,m=(2q^2+3q+1)

hence, square of any positive odd integer is of the form 8m+1