Very nice one, anybody has a solution?

I want to go back to this nice problem, but I don't know any other means but to answer this post and then click on "View your posts" in order to go back. Is there other ways (such as a private watch list)? I will try to solve this nice problem.

it's not a solution, but i think that it can gives some thing we take $ P,Q\in\mathbb{Z}$ such $ P(k+1)-P(k)=k^{n},Q(k+1)-Q(k)=k^{2n}$ let $ S(x,y)=P(x)-Q(y)\in\mathbb{Z}[X,Y]$ we must prove that $ S(x+m,y+m)=S(x,y)$ have infnit solution in $ \mathbb{N}^2$

I think we could solve this problem using the result ( [(x+1)^n+....+(x+m)^n ]/n)^(1/n) - x is bounded as x tends to infinity, that help us to show that |x-y^2| must be bounded....

mto wrote: I think we could solve this problem using the result ( [(x+1)^n+....+(x+m)^n ]/n)^(1/n) - x is bounded as x tends to infinity why $ (\frac {(x + 1)^n + .... + (x + m)^n }{n})^{\frac {1}{n}} - x$ is bounded?it is not...

Hint: a) Show that there exist a constant $ A$ such that $ x\geq y^2+(m+1)y+A$. b) Show that there exist a constant $ B$ such that the equation has only finitely many solutions $ x,y$ satisfying $ x\geq y^2+(m+1)y+B$. More hint here: calculate the terms involving $ y^{2n-2}$ in both sides of the inequality.

Can anyone explain how to use this hint ?

We claim that for $y$ sufficiently large there exists a positive integer $C$ such that \[y^2 + (m + 1)y - C \le y^2 + (m + 1)y + C\] Notice that the problem statement follows immediately; for any particular $y$, the LHS is monotonically increasing so there can be at most one solution for $x$ and thus finitely many for all $y$ smaller than our threshold. Any other solution pairs $(x, y)$ must satisfy one of \[x = y^2 + (m + 1)y - C\] \[x = y^2 + (m + 1)y - C + 1\] \[\dots\] \[x = y^2 + (m + 1)y + C\] Substituting into the given equation, each gives a degree $2n$ polynomial in $y$, so in all these can give at most $(2C + 1)(2n)$ (a finite number of) solutions, which when added to our previously finite count of solutions remains finite. Plug in $x = y^2 + (m + 1)y \pm C$ and examine the coefficients of the highest powers of $y$ on either side. The coefficient of $y^{2n}$ in the left and right hand sides are both $m$. Similarly, the coefficient of $y^{2n - 1}$ in the left and right hand sides are each $nm(m + 1)$. Now consider $[y^{2n - 2}]$. In the left hand side, this is \[m(m + 1)^2\binom{n}{2} + \frac{nm(m + 1)}{2} \pm m\binom{n}{1}C \; \; \; \; \; \; \; \; \; (1)\] while in the right hand side it is \[\frac{m(m + 1)(2m + 1)}{6}\binom{2n}{2} \; \; \; \; \; \; \; \; \; \; (2)\] It is easy to see we can choose and integer $N$ such that for all $C \ge N$, $(1)$ exceeds $(2)$; similarly, we can choose $M$ such that for all $C \le M$, $(2)$ exceeds $(1)$. Now just choose $C = \max(|M|, |N|)$. For $y$ sufficiently large, a polynomial in $y$ behaves like its leading term. In particular, the polynomial $LHS - RHS$ behaves like its leading term \[ \left( \left(m(m + 1)^2\binom{n}{2} + \frac{nm(m + 1)}{2} \pm m\binom{n}{1}C \right) - \left( \frac{m(m + 1)(2m + 1)}{6}\binom{2n}{2} \right) \right)y^{2n - 2}\] which grows without bound when the coefficient is non-zero. Thus for $y$ sufficiently large, all solutions $(x, y)$ to the equation must satisfy \[y^2 + (m + 1)y - C \le x \le y^2 + (m + 1)y + C\] for the chosen positive integer $C$, and we are done.