Question

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

plz answer fast.

Answer 1

Let 2k+1 be any odd integer.

Now, (2k+1)^2
= 4k^2 + 4k + 1
= 4k(k+1) + 1

Now, k and k+1 are consecutive integers. So one of them must be even. So, k(k+1) is divisible by 2.

Let k(k+1) = 2q

So,
4k(k+1) +1
= 8q +1

Thus, square of any odd positive integer is of the form 8q + 1