Question

Show that the square of any +ve odd integer is of the form 8m+1 for some integer m

Answer 1

a=8q+r
r=1,2,3,4,5,6,7,0
At r=0
a^2=8q^2 =64q square =8(8q) =8m where m = 8q even
r=1
a^2=(8q+1)^2=64q sq.+16q+1=8 (8q+2)+1= 8m+1 where m = 8q+2
So square of any odd integer in the form 8q+1

Answer 2

let x be the positive integer
x=8m+1
x²=(8m+1)²
x²=(8m)²+2*8m*(1)+1
x²=64m²+16m+1
x²=8(8m²+2m)+1
x²=8m+1 [ let 8m²+2m =m]
∴ proved