# solutions to $f(y) = n(n + 1) \ldots (n + m - 1)$

I was reading a paper about solutions to $f(y) = P(m)$, where $f(y) \in \mathbb Z[y]$ and $P(m) = n(n + 1) \ldots (n + m - 1)$ is a product of $m$ consecutive integers.

On the first page, it is mentioned that it is not difficult to get the solutions in the case that $f(y)$ is irreducible. However, I was confused about two things.

• First, the author mentions a condition on $f(y)$ which implies that the equation $f(y) = P(m)$ has no solution, but doesn't mention when a solution does exist.
• Secondly, I'm not sure how to generalize the method used in the example in the paper. In the example given, the number of cases where it is possible to have a solution is quite small, so it isn't hard to check what the solutions are (if any).
I think this might be a problem when $f(y)$ has large degree. Also, I tried to solve $f(y) = P(m)$ when $f(y) = py + 1$, where $p$ is an odd prime. Using the condition given, I found that the smallest prime $q$ which does not divide $f(y)$ for any $y$ is $p$. So, there are no solutions when $m > p$. But I'm not sure how to start finding general solutions (or finding conditions for $m$ when solutions do exist).

Is there something I'm misinterpreting in the paper? Any suggestions on how to start or other references to look at?

Note: This is cross-posted from math.SE: https://math.stackexchange.com/questions/419918/solutions-to-fy-nn-1-ldots-n-m-1

As to your first question, indeed, it is hard to construct explicit example, when solution exists unless $P(m)=f(y)$ is an identity. When the author claims, that for an irreducible polynomial it is easy to solve the problem, most likely, he intended to write that it is easy to get the condition on $m,$ when solution does not exist.
Indeed, if $f(y)$ has large degree, then the smallest $p$ that does not divide $f(y)$ might be large and you will have to consider many cases for $m.$ So this algorithm is far from being effective in this particular case. Finally, for $py+1=P(m)$ you need to find such $m$ that $P(m)=1(\mod p)$ for some $n.$ In this case, you can take $m=p-1$ and $n=1(\mod p)$ to get infinite number of solutions (just ue Willson's theorem). For each particular $m,$ you will have to check the values of $P(m)$ at $n=1,$ $n=2,$... $n=p-m$ and find out where the congruence has solution.
As to the references, I would suggest to look at the paper Berend and Harmse http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/home.html where they consider equation $f(y)=m!$ and prove that it has finitely many solutions for large class of polynomials $f(y).$