Is there a classification of all the couples of positive integers $(n,m)$ such that $m\mid 2n$ and $4n\mid m^2$? The couples are infinite and I think there is no bound on $m$. I want to know because I want to classify isotropic subgroups of the discriminant group of an even lattice of rank $1$.
Thank you very much in advance.
Things are actually quite straightforward. Assuming that $m, n \in \mathbb{N}$ are both natural and that $4n|m^2$ we infer in particular that $2|4n|m^2$ whence by transitivity $2|m^2$ and thus $2|m$. This means we can write $m=2r$ for a certain $r \in \mathbb{N}$.
Turning now to the first divisibility relation we gather from $2r|2n$ that $r|n$, whereas the second divisibility relation reduces to $n|r^2$. This means that we have the expression $n=kr$, with $k \in \mathbb{N}$ and such that $k|r$ if $r \neq 0$.
To conclude, apart from the trivial pair $(m, n)=(0, 0)$ we have in general $(m, n)=(2r, kr)$, where $r \in \mathbb{N}^{\times}$ is arbitrary and $k|r$ is a divisor of $r$.