For every non-negative integer z, z2 - 3z is an even integer. Prove this statement. So far, I have learned about direct proofs and indirect proofs such as contraposition and contradiction.
An integer z is odd when there is an integer a so that z = 2a+1. An integer is even when there is an integer a such that z = 2a.
The Attempt at a Solution
I broke it down into to two parts where x is even and another where x is odd:
For even, I substituted z = 2a into z2 - 3z to get (2a)2 - 3(2a). After expanding, I got 4a2 - 6a and further simplified it to get 2a(2a - 3).
I tried using proof by contradiction by taking the negation of the statement and saying suppose there is at least one non-negative integer z, z2 - 3z is an odd integer.
For odd, I substituted z = 2a + 1 to get (2a+1)2 - 3(2a + 1). After expanding, I got (4a2 + 4a + 1) - 6a - 3. Then, 4a2 - 2a - 2. Finally, I simplified it to get 2(2a2 - a) - 2. (2a2 - a) is an integer because it is the difference of two integers. However, this does not match up with the definition of odd, which is z = 2a + 1. This means that there is not at least one non-negative integer z, z2 - 3z is an odd integer. Therefore, the statement that for every non-negative integer z, z2 - 3z is an even integer is true.
I got confused somewhere down the line and I don't think that the solution I came up with is correct. I would really appreciate some assistance with this question.