Question

Show that the square of an odd positive integer is of the form 8q+1 , for some integer q

Answer 1

Answer:

Let n be an odd integer (it's not necessary for it to be positive!).

Then n = 2k + 1 for some integer k.

Thus

n² = ( 2k + 1 )²

   = 4k² + 4k + 1

   = 4k(k + 1) + 1

Since k and k + 1 are consecutive integers, one of them is even.  Therefore 4k(k + 1) is a multiple of 8.  That it, 4k(k + 1) = 8q for some integer q.  Hence

n² = 8q + 1.