Question

Show that p square will leaves a remainder 1 when divided by 8 if P is any positive odd integer

Answer 1

a=bq+r
let b=8
possible values of r are = 0,1,2,3,4,5,6,7
let p=a
p=8q+r
squaring on both sides
p^2=64q^2+16qr+r^2
p^2=8(8q^2+2qr)+r^2
let 8q^2+2qr be any integer n
p^2=8n+r^2
(p^2 is of when p is of because odd multiplied by of is odd. but 8n is even. so remainder is always odd)
possible values of r=1,3,5,7
if r=1
p^2=8n+1
leaves remainder 1
if r=3
p^2=8n+9=8(n+1)+1=8q+1
leaves remainder 1

similarly substitute ask remaining basis of r and then take 8 common you will get remainder 1

hope it helps you
pls mark brainliest