Proof Problem - 1

An odd integer has the form of $\large 2m +1$ , in other words $\large even integer + 1 = odd integer$ .

$\large (2m + 1)^2 = (2m + 1)(2m + 1) = {4m}^2 + 4m + 1$

$\large {4m}^2 + 4m + 1 = 2({2m}^2 + 2m) + 1$

Using the first statement, it is proven that the square of an odd integer is odd

In order for a number to be even, it must contain a 2 in its prime factorization. An odd number doesn't contain any 2s in its prime factorization, so an odd number squared won't contain any 2s in its prime factorization, so it will be even.