This is probably of limited help in general, but if you write the second and third row of the matrix in MHMertens's answer as

$$\begin{pmatrix}{f_0}&{f_1}&{f_2}&{\cdots}\cr {g_0}&{g_1}&{g_2}&{\cdots}\cr\end{pmatrix}$$

where the second row is clearly the Fibonacci sequence, then the third row satisfies the recursion

$$g_{n+1} = 2f_{n+1}f_n - g_n$$

and therefore consists of all integers.

The real thing to prove is that $g_n+g_{n-1}=2f_nf_{n-1}$ implies $g_{n+1}+g_n = 2f_{n+1}f_n$, and this is easy since the Fibonacci recursion $f_n+f_{n-1}=f_n$ gives

$$g_{n+1}+g_n = {f_{n+1}g_{n-1}+f_ng_n\over f_{n-1}}+g_n = {f_{n+1}(g_{n-1}+g_n)\over f_{n-1}}.$$