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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 $b$?

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    $\begingroup$ 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.) $\endgroup$ – LeechLattice Jun 28 at 10:07
  • $\begingroup$ 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). $\endgroup$ – LeechLattice Jun 28 at 12:02
  • $\begingroup$ @LeechLattice Thanks, I edited the question. $\endgroup$ – jack yesterday

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