Prove that if x^2 is odd then x is odd. (Direct proof)

Assume x is an integer. If x^2 (x squared) is odd then x is odd.

First lets look at some examples

3^2 is 9
5^2 is 25
7^2 is 49
9^2 is 81

So it does appear that for any odd integer, its square is also an integer, how can we prove this for all cases?

First we assume x is odd, then x=2y+1 for some integer y.

So now x^2 = (2y+1)(2y+1) = 4y^2 + 4y + 1= 2(2y^2 + 2) + 1 which is odd.

Prove that if x is even and y is odd, then x+y is odd

Let x and y be integers. Prove that if x is even and y is odd, then x+y is odd.

x is even, therefore x=2k for some integer k.
y is odd, therefore y=2j+1 for some integer j.

by substitution into x+y=odd we get

2k + 2j + 1 = odd

2k + 2j + 1 = 2(k+j) + 1, which is odd.