# Is it possible to determine if $\frac{xy^2}{2}$ is an even number?

#### The problem

Given $$x, y \in \mathbb{Z}$$, is it possible to determine if $$\frac{xy^2}{2}$$ is an even number?
$$x$$ and $$y$$ are consecutive numbers and $$x$$ is even.

#### My attempt

Assuming $$n$$ is an integer, it can be rewritten as
$$\frac{2n\times(2n-1)}{2} = n\times(2n-1)\times(2n-1)$$ Where $$(2n-1)$$ is always odd, so it all comes down to $$n$$, which is unknown. However, the tutor said it can be determined for sure. Another way of going about this would be to say $$x$$ is even and $$y$$ is odd, so $$xy$$ is even, therefore $$xy^2$$ is even as well.
What can be said about $$\frac{xy^2}{2}$$ though?

• Nothing can be said about it. Your tutor is incorrect. It depends on $n$. Or, if you want to look at it this way, $xy^2$ is even, but $\frac{\text{even}}{2}$ can be even or odd. $\frac{6}{2}$ is odd, but $\frac{8}{2}$ is even. Jul 19, 2021 at 15:47
• You're right, the answer depends on the parity of $n.$ Jul 19, 2021 at 15:48
• @Luna145 thank you. Now it seems to me that he might have meant $yx^2$ instead. Jul 19, 2021 at 15:51

Well $$\frac{xy^{2}}{2}$$ is only even if $$xy^{2}$$ has a factor of $$4 = 2^2$$

Therefore, $$\frac{xy^2}{2}$$ is even always if

$$y$$ is divisible by two

or

$$x$$ is divisible by four
However $$x$$ is even so $$y$$ cannot be divisible by two since they are consecutive. Therefore $$x$$ must be divisible by four.

We have to determine if $$\dfrac{xy^2}{2}$$ is even

So using common sense, a even number can be shown as $$2k$$

$$xy^2=4k$$

Case 1: $$x$$ must be divisible by $$4$$. True enough no counterparts

Case 2: $$y$$ is divisible by $$2$$ True enough

Case 3: $$x,y$$ both are even. This case is broken as you have proved $$x$$ and $$y$$ are composite so both can't be even at same time