Given two irrational numbers $a,b\in\mathbb R\setminus\mathbb Q$ how many possible solutions exist to the equation
$$an+b=m$$
where $n,m\in \mathbb Z$.
For example, take $a=\sqrt 2$ and $b=-\sqrt 2$. Then clearly we have solutions at $(n=1,m=0)$.
If $b=ta$ for some $t\in\mathbb Z$ then we have
$$an+ta=a(t+n)=m$$
and since the LHS must be irrational for all $n\ne t$ we can say that there exists only one unique solution at $(n=-t,m=0)$.
Is there some general way to count the number of allowed combinations of $n,m$ for all possible $a,b$? My conjecture is that there is $1$ allowed solution to the equation for each and any $a,b$.
There is at most one solution. Suppose $m_1-an_1=m_2-an_2$, then we get $m_1-m_2=a(n_1-n_2)$. The RHS is irrational if $n_1\ne n_2$, the LHS is integral. Thus we get $n_1=n_2,m_1=m_2$. We may have zero solutions, like $a=\sqrt3,b=\pi$.
Note that we get a solution iff $\exists n,m\in\Bbb Z|b=m+an$, i.e. $b\in\Bbb Z[a]$.