TITLE: Proving that every term of the sequence is an integer
QUESTION [16 upvotes]: Let $m,n$ be nonnegative integers.
The sequence $\{a_{m,n}\}$ satisfies the following three conditions.

For any $m$, $a_{m,0}=a_{m,1}=1$
For any $n$, $a_{0,n}=1$
For any $m\ge0, n\ge1$, $a_{m+1,n+1}a_{m,n-1}=a_{m+1,n-1}a_{m,n+1}+a_{m+1,n}a_{m,n}$

Prove that $a_{m,n}$ is an integer for any $m\ge0, n\ge0$.
I don't have any good idea. I need your help.

REPLY [8 votes]: This looks like it might be a "rank two elliptic divisibility sequence", i.e., let $E$ be an elliptic curve, let $P$ and $Q$ be independent points in $E(\mathbb{Q})$, and write
$$
  x(mP+nQ) = A(m,n)/D(m,n)^2.
$$
Then the doubly-indexed sequence $D(m,n)$ is a rank two elliptic divisibility sequence, or as they were named by Kate Stange, an elliptic divisibility net. Kate made a detailed study of these sequences in her thesis (Algebra and Number Theory, 5-2 (2011), 197-229; it's also available on the ArXiv at http://arxiv.org/abs/0710.1316), including a description of the recurrence relation(s) that they satisfy. 
Anyway, I don't know for sure if your rank two sequence fits into Kate's elliptic net framework, but it's quite possible that it does.
(I'm actually cheating a little bit here, one really needs to use division polynomials instead of writing $x(mP+nQ)$ as a fraction in case there's cancellation between the numerator and the denominator. But this will only affect primes of bad reduction.)