Proof That The Result of This Function is An Even Number?

Question

I feel like the solution should be obvious, but I can't figure out how to formally prove it.

I have a relation of $(a,b)$ pairs such that $3a - b^2$ is an even integer. I want to know if for an $(a,a)$ pair (ie. two of the same number), the result is even.

I know that obviously even numbers divide by two, and that a number divides by $2$ if it is some multiple of $2$, i.e. $x = 2k$. Plugging in, I have $3a - a^2$, which I can re-write as $a(3-a)$. Just by trying a few test integers ($1$, $2$, etc.) it seems to hold, but I'd rather do it without an inductive proof since I'm sure it is possible.

Any pointers in the right direction? Thanks!

Answer 1

Hint: observe two cases:

1. $a = 2k$ for some $k \in \mathbb{N}$ (i.e., take an even $a$);
2. $a = 2k-1$ for some $k \in \mathbb{N}$ (i.e., take an odd $a$).

Answer 2

You have that $3a-a^2 = a(3-a)$, which is true.

If $a$ is even, then the result must be even, because multiplication by an even number always yields an even number.

If $a$ is odd, then $3-a$ is even. This is easy to see by examining that $3-a = 2 + (1-a)$. Since $a$ is odd, so is $-a$, and by adding $1$ to $-a$, you must get an even number.

Answer 3

You can also use

$$3a-a^2=2-(2-3a+a^2)=2-(a-1)(a-2)$$

$(a-1)(a-2)$ is even as the product of two consecutive integers, and $2$ is even.