Question

show that the square of any positive odd integer is of the form 8m + 1, for some integer m

Answer 1

Answer:

let the square of any positive integer be a

a = 4m

b= 4

we know r can be 0,1 or 2

as b> r

so taking a=4m +1

a² = (4m +1)²

a = 16 m + 1 + 2*16m

a² = 16m +1+32m

taking 8 out as common

a² = 8(2m + 4m)+1

let 2m + 4m = m

a² = 8m +1

Proved