Prove that the elements of the sequence are integers A sequence is given by the following recursion. Prove that the elements of this sequence are integers!
$a_0=1$
$a_1=41$
$a_{n+2}=3a_n+\sqrt{8(a_n^2+a_{n+1}^2)}$

Sequence is Integral
Suppose
$$
a_{n+2}=3a_n+b_n\tag{1}
$$
where
$$
b_n^2=8\!\left(a_n^2+a_{n+1}^2\right)\tag{2}
$$
Then
$$
\begin{align}
b_{n+1}^2
&=8\!\left(\color{#C00}{a_{n+2}^2}+\color{#090}{a_{n+1}^2}\right)\\
&=8\!\left(\color{#C00}{9a_n^2+6a_nb_n+b_n^2}+\color{#090}{\frac{b_n^2}8-a_n^2}\right)\\
&=64a_n^2+48a_nb_n+9b_n^2\\[6pt]
&=(8a_n+3b_n)^2\tag{3}
\end{align}
$$
Therefore,
$$
b_{n+1}=8a_n+3b_n\tag{4}
$$
We can continue the integer sequences using $(1)$ and $(4)$:
$$
\begin{bmatrix}a_{n+1}\\a_{n+2}\\b_{n+1}\end{bmatrix}
=
\begin{bmatrix}0&1&0\\3&0&1\\8&0&3\end{bmatrix}
\begin{bmatrix}a_n\\a_{n+1}\\b_n\end{bmatrix}\tag{5}
$$
That is,
$$
\begin{bmatrix}a_n\\a_{n+1}\\b_n\end{bmatrix}
=
\begin{bmatrix}0&1&0\\3&0&1\\8&0&3\end{bmatrix}^{\,\large n}
\begin{bmatrix}1\\41\\116\end{bmatrix}\tag{6}
$$

Closed Form for the Sequence
Using the Jordan Normal Form of the matrix in $(6)$, we get
$$
\begin{bmatrix}a_n\\a_{n+1}\\b_n\end{bmatrix}
=
\begin{bmatrix}
-1&\frac{-1-\sqrt3}8&\frac{-1+\sqrt3}8\\
1&\frac{1-\sqrt3}8&\frac{1+\sqrt3}8\\
2&1&1
\end{bmatrix}
\begin{bmatrix}
-1&0&0\vphantom{\frac{\sqrt3}8}\\
0&2-\sqrt3&0\vphantom{\frac{\sqrt3}8}\\
0&0&2+\sqrt3
\end{bmatrix}^{\,\large n}
\begin{bmatrix}
\frac{22}3\\
\frac{152-84\sqrt3}3\\
\frac{152+84\sqrt3}3
\end{bmatrix}\tag{7}
$$
from which, we can extract
$$
a_n=\frac{25+17\sqrt3}6\left(2+\sqrt3\right)^n-\frac{22}3(-1)^n+\frac{25-17\sqrt3}6\left(2-\sqrt3\right)^n\tag{8}
$$
It is not immediately evident that $(8)$ is always an integer; however, since we have $\left(x-2-\sqrt3\right)\left(x-2+\sqrt3\right)(x+1)=x^3-3x^2-3x+1$, equation $(8)$ implies that
$$
a_n=3a_{n-1}+3a_{n-2}-a_{n-3}\tag{9}
$$
which does imply that $a_n$ is always an integer since $a_0=1$, $a_1=41$, and $a_2=119$.
