A "number triangle" $(t_{n, k})$ $(0 \le k \le n)$ is defined by $t_{n,0} = t_{n,n} = 1$ $(n \ge 0),$ $$t_{n+1,m} =(2 -\sqrt{3})^mt_{n,m} +(2 +\sqrt{3})^{n-m+1}t_{n,m-1} \quad (1 \le m \le n).$$ Prove that all $t_{n,m}$ are integers.
I find it hard to prove that they are all integers because from the table below there doesn't seem to be a pattern. I think induction might work if we use strong induction on $m$ for a fixed $n$. We know that it holds for $t_{n,0}$ so we have the base case. How should I write the inductive proof?
Assume that $t_{n,k}$ are integers for all $n < r$. Now, we have $$t_{r,1} = (2-\sqrt{3})t_{r-1,1}+(2+\sqrt{3})^{r-1}t_{r-1,0} = (2-\sqrt{3})t_{r-1,1}+(2+\sqrt{3})^{r-1}.$$ Then we have $t_{r-1,1} = (2-\sqrt{3})t_{r-2,1}+(2+\sqrt{3})^{r-2}t_{r-2,0} = (2-\sqrt{3})t_{r-2}+(2+\sqrt{3})^{r-2}$, which we know is an integer, but I don't see where to go from here.
Here is a table I made:
$t_{0,0} = 1$
$t_{1,0} = 1$
$t_{1,1} = 1$
$t_{2,0} = 1$
$t_{2,1} = (2-\sqrt{3})+(2+\sqrt{3}) = 4$
$t_{2,2} = 1$
$t_{3,0} = 1$
$t_{3,1} = (2-\sqrt{3})t_{2,1}+(2+\sqrt{3})^2 t_{2,0} = 4(2-\sqrt{3})+(2+\sqrt{3})^2 = 15$
$t_{3,2} = (2-\sqrt{3})^2 t_{2,2}+(2+\sqrt{3}) t_{2,1} = (2-\sqrt{3})^2+4(2+\sqrt{3}) = 15$
$t_{3,3} = 1$
$t_{4,0} = 1$
$t_{4,1} = (2-\sqrt{3}) t_{3,1}+(2+\sqrt{3})^3 t_{3,0} = 15(2-\sqrt{3})+(2+\sqrt{3})^3 = 56$
$t_{4,2} = (2-\sqrt{3})^2 t_{3,2}+(2+\sqrt{3})^2t_{3,1}= 15(2-\sqrt{3})^2 +15(2+\sqrt{3})^2 = 210$
$t_{4,3}= (2-\sqrt{3})^3 t_{3,3}+(2+\sqrt{3})t_{3,2} = (2-\sqrt{3})^3+15(2+\sqrt{3}) = 56$
$t_{4,4} = 1$
$t_{5,0} = 1$
$t_{5,1} = (2-\sqrt{3})t_{4,1}+(2+\sqrt{3})^4 t_{4,0} = 56(2-\sqrt{3})+(2+\sqrt{3})^4 = 209$
$t_{5,2} = (2-\sqrt{3})^2 t_{4,2}+(2+\sqrt{3})^4 t_{4,1} = 210(2-\sqrt{3})^2+56(2+\sqrt{3})^3 = 2926$
$t_{5,3} = (2-\sqrt{3})^3 t_{4,3}+(2+\sqrt{3})^2 t_{4,2} = 56(2-\sqrt{3})^3+210(2+\sqrt{3})^2 = 2926$
$t_{5,4} = (2-\sqrt{3})^4 t_{4,4}+(2+\sqrt{3}) t_{4,3} = (2-\sqrt{3})^4+56(2+\sqrt{3})^2 = 209$
$t_{5,5} = 1$
$t_{6,0} = 1$
$t_{6,1} = (2-\sqrt{3})t_{5,1}+(2+\sqrt{3})^{5}t_{5,0} = 209(2-\sqrt{3})+(2+\sqrt{3})^5 = 780$
$t_{6,2} = (2-\sqrt{3})^2 t_{5,2}+(2+\sqrt{3})^4 t_{5,1} = 2926(2-\sqrt{3})^2+209(2+\sqrt{3})^4 = 40755$
$t_{6,3} = (2-\sqrt{3})^3 t_{5,3}+(2+\sqrt{3})^3 t_{5,2} = 2926(2-\sqrt{3})^3+2926(2+\sqrt{3})^3 = 152152$
$t_{6,4} = (2-\sqrt{3})^4 t_{5,4}+(2+\sqrt{3})^{2}t_{5,3} = 209(2-\sqrt{3})^4+2926(2+\sqrt{3})^3 = 40755$
$t_{6,5} = (2-\sqrt{3})^5 t_{5,5}+(2+\sqrt{3})t_{5,4} = (2-\sqrt{3})^5+209(2+\sqrt{3}) = 780$
$t_{6,6} = 1$