Might this help. Let ${b}_{m,n}=\begin{pmatrix} {a}_{m,n} &{a}_{m,n+1} &{a}_{m,n+2} \\ {a}_{m+1,n} &{a}_{m+1,n+1} &{a}_{m+1,n+2} \end{pmatrix}$

Your iteration formula says that if you multiply the top left and bottom right corner numbers of ${b}_{m,n}$, subtract the product of the top right and bottom left corner numbers, then you get the product of the top and bottom middle numbers. The first two rows of your numbers are known to be integers (Dietrich), as are the first three columns (J.J). Hence ${b}_{0,n}$ and ${b}_{m,0}$ contain integers. Now consider the sequence of matrices ${b}_{0,0}$, ${b}_{0,1}$, ${b}_{1,0}$, ${b}_{0,2}$, ${b}_{1,1}$, ${b}_{2,0}$ etc. The only unknown number in ${b}_{1,1}$ is the bottom right hand number and this is true of each of these 2 by 3 matrices when we leave the first two rows or first 3 columns. Could that form the basis of a proof by induction?

This might help. Let $${b}_{m,n}=\begin{pmatrix} {a}_{m,n} &{a}_{m,n+1} &{a}_{m,n+2} \\ {a}_{m+1,n} &{a}_{m+1,n+1} &{a}_{m+1,n+2} \end{pmatrix}$$

Your iteration formula says that if you multiply the top left and bottom right corner numbers of ${b}_{m,n}$, subtract the product of the top right and bottom left corner numbers, then you get the product of the top and bottom middle numbers. The first two rows of your numbers are known to be integers (Dietrich), as are the first three columns (J.J). Hence ${b}_{0,n}$ and ${b}_{m,0}$ contain integers. Now consider the sequence of matrices ${b}_{0,0}$, ${b}_{0,1}$, ${b}_{1,0}$, ${b}_{0,2}$, ${b}_{1,1}$, ${b}_{2,0}$ etc. The only unknown number in ${b}_{1,1}$ is the bottom right hand number and this is true of each of these $2 \times 3$ matrices when we leave the first two rows or first 3 columns. Could that form the basis of a proof by induction?

Might this help. Let $${b}_{m,n}=\begin{pmatrix} {a}_{m,n} &{a}_{m,n+1} &{a}_{m,n+2} \\ {a}_{m+1,n} &{a}_{m+1,n+1} &{a}_{m+1,n+2} \end{pmatrix}$$

Your iteration formula says that if you multiply the top left and bottom right corner numbers of ${b}_{m,n}$, subtract the product of the top right and bottom left corner numbers, then you get the product of the top and bottom middle numbers. The first two rows of your numbers are known to be integers (Dietrich), as are the first three columns (J.J). Hence ${b}_{0,n}$ and ${b}_{m,0}$ contain integers. Now consider the sequence of matrices ${b}_{0,0}$, ${b}_{0,1}$, ${b}_{1,0}$, ${b}_{0,2}$, ${b}_{1,1}$, ${b}_{2,0}$ etc. The only unknown number in ${b}_{1,1}$ is the bottom right hand number and this is true of each of these 2 by 3 matrices when we leave the first two rows or first 3 columns. Could that form the basis of a proof by induction?

Might this help. Let ${b}_{m,n}=\begin{pmatrix} {a}_{m,n} &{a}_{m,n+1} &{a}_{m,n+2} \\ {a}_{m+1,n} &{a}_{m+1,n+1} &{a}_{m+1,n+2} \end{pmatrix}$

Your iteration formula says that if you multiply the top left and bottom right corner numbers of ${b}_{m,n}$, subtract the product of the top right and bottom left corner numbers, then you get the product of the top and bottom middle numbers. The first two rows of your numbers are known to be integers (Dietrich), as are the first three columns (J.J). Hence ${b}_{0,n}$ and ${b}_{m,0}$ contain integers. Now consider the sequence of matrices ${b}_{0,0}$, ${b}_{0,1}$, ${b}_{1,0}$, ${b}_{0,2}$, ${b}_{1,1}$, ${b}_{2,0}$ etc. The only unknown number in ${b}_{1,1}$ is the bottom right hand number and this is true of each of these 2 by 3 matrices when we leave the first two rows or first 3 columns. Could that form the basis of a proof by induction?