TITLE: Irrationality of $\sqrt{2m}$
QUESTION [6 upvotes]: I have this problem:

If $\sqrt{m}$ is irrational, and $m$ is odd, so:
Is $\sqrt{2m}$ always irrational ?

My thought was:
I have a $m$ number, that can be of the form $(2n \pm 1)$, that is square is irrational.
To find a counterexample, I need this:
$2m = x^2$,
then I need an integer, that have perfect square, divided in $2$ is an odd number, that is equal to:
Lets $x$ any integer:
$\frac{x^2}{2} = k$
Then, $k$ must be an integer, since it is divisible by 2 and must be a number of form $(2n \pm1)$  that is
$m$.
With these conclusions, I can reform my problem to:

Is there an perfect square, which is divisible by $2$, and this
division is an odd number?

I do not ask this directly, because there may be an easier approach, but still this form seems very interesting to me.
PS: I have tested it in programming between $[1, 1000]$ and there is no perfect square, with these characteristics.

REPLY [4 votes]: That is equivalent to: There are no square numbers of the form $2m$, where $m$ is odd.
Proof: Take any number $x$, if it is odd then $x^2$ is odd. If it is even, then $x^2$ is divisible by $4$. That means that if $2m = x^2$, $m$ must be divisible by $2$, which contradicts the statement $m$ is odd.