# $2n^2-\lfloor m^b\rfloor=k$ has only finitely many integer solutions

Let $k$ be an nonzero integer and $b>2$ a real. Is it true that there exist only finitely many positive integer pairs $(n,m)$ for which $$2n^2-\lfloor m^b\rfloor=k?$$

I don't know the answer, but I guess it is positive.. To be precise, I think the following may be true:

Conjecture: Let $\alpha,\beta$ be positive integers, $k$ a nonzero integer, and $a,b$ be distinct reals greater than $1$. Then there exist only finitely many positive integer pairs $(n,m)$ for which $$\alpha \lfloor n^a\rfloor-\beta\lfloor m^b\rfloor=k.$$ [It includes, as a special case, this other question (which has a positive answer).]

• This is an interesting question. Where did you encounter it? – abiessu Sep 27 '15 at 1:54
• I was just trying to construct an example related to this MO question mathoverflow.net/questions/219274/… . Anyway, it recalled me a well-known problem related to Mihailescu theorem, that's why I posed it here :) – Paolo Leonetti Sep 27 '15 at 7:59

## 1 Answer

No, it is not. Start with $2\cdot 1^2-[1^3]=1$ and notice that this equation holds for $b\in I_1$ where $I_1$ is some closed interval starting at $3$. Now suppose that we have already several solutions of $2n^2-[m^b]=1$ with the possibility to take any $b\in I$ in the equations. Notice that the ratio of two consecutive numbers of the kind $2n^2-1$ tends to $1$ when we go to infinity while for any $a<c$ with $a,c\in I$ we have $m^c/m^a\to \infty$ as $m\to\infty$. So, by continuity, for a sufficiently large $m$, we can find $B\in(a,c)$ such that $m^B=2n^2-1$ for some integer $n$. Then we will have $2n^2-[m^b]=1$ for all $b$ in some short closed interval $J\subset I$ starting at $B$. This process will create a nested sequence of closed intervals, and their common point $b$ will be a counterexample to the conjecture.

• Very nice fedja! – Paolo Leonetti Aug 25 '16 at 8:10