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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Moni145
    Jul 19, 2021 at 15:47
  • 1
    $\begingroup$ You're right, the answer depends on the parity of $n.$ $\endgroup$
    – user376343
    Jul 19, 2021 at 15:48
  • $\begingroup$ @Luna145 thank you. Now it seems to me that he might have meant $yx^2$ instead. $\endgroup$ Jul 19, 2021 at 15:51

2 Answers 2


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


$x$ 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 $2k$


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


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