# Infinitely many $n$ such that $\gcd(\lfloor n\sqrt{2}\rfloor, \lfloor n\sqrt{3}\rfloor)=m$

Is it true that for any positive integer $$m$$ there are infinitely many positive integers $$n$$ such that $$\gcd(\lfloor n\sqrt{2}\rfloor, \lfloor n\sqrt{3}\rfloor)=m$$?

$$\lfloor x \rfloor$$ is the floor function of $$x$$ and $$\gcd$$ is the greatest common divisor.

In fact we can prove that this result is true for $$\gcd(\lfloor n\sqrt{2}\rfloor, n)=m$$ (consider a solution $$(x_k,y_k)$$ to the Pell equation $$x^2-2y^2=-1$$, such that $$x_k>m$$ and take $$n=my_k$$).

Does this result hold if instead of $$\sqrt{3}$$ we have $$\sqrt{k}$$, where $$k$$ is not a perfect square, $$k>2$$?

Can this result be generalized in any way?
If $$a$$ and $$b$$ are positive real numbers such that $$\gcd(\lfloor na \rfloor$$, $$\lfloor nb \rfloor)=m$$ holds for infinitely many positive integers $$n$$ for each $$m$$, what can be said about $$a$$ and $$b$$?

• By the multidimensional equidistribution theorem, for any $m$, the probability of $m$ dividing both $\lfloor{n\sqrt 2}\rfloor$ and $\lfloor{n\sqrt 3}\rfloor$ is $1/m^2$. $gcd(\lfloor{n\sqrt 2}\rfloor, \lfloor{n\sqrt 3}\rfloor)=m$ is equivalent to $m$ dividing both $\lfloor{n\sqrt 2}\rfloor$ and $\lfloor{n\sqrt 3}\rfloor$, but neither of $2m,3m,5m,7m,11m,...$ divides both $\lfloor{n\sqrt 2}\rfloor$ and $\lfloor{n\sqrt 3}\rfloor$. (One should use an effective estimate here, in order to deal with infinite number of primes.) – LeechLattice Jun 28 at 10:07
• The answers to your last two questions are both "no". The correct generalization is requiring $a,b$ nonzero and $a/b$ irrational. If $a/b$ is rational, let $a/b=p/q$ ($p,q\in \mathbb Z$), and $q\lfloor na \rfloor-p \lfloor nb \rfloor$ is bounded independent of $n$, so $\gcd(\lfloor na \rfloor,\lfloor nb \rfloor)$ is either bounded or of the order of $n$. If $a/b$ is irrational, the equidistribution theorem works (use one-dimensional equidistribution if only one of $a$ and $b$ is irrational). – LeechLattice Jun 28 at 12:02
• @LeechLattice Thanks, I edited the question. – jack yesterday