Solution:

Let $A = \sqrt{5} + 2$ and $B = \sqrt{5} - 2$. Then $a_k = A^k + B^k$.

Note that $A \cdot B = (\sqrt{5} + 2)(\sqrt{5} - 2) = (\sqrt{5})^2 - 2^2 = 5 - 4 = 1$.

Also, $A + B = (\sqrt{5} + 2) + (\sqrt{5} - 2) = 2\sqrt{5}$.

We want to show that $a_{m+n} + a_{m-n} = a_m a_n$.

Compute $a_m a_n$:

$$a_m a_n = (A^m + B^m)(A^n + B^n) = A^{m+n} + A^m B^n + B^m A^n + B^{m+n}$$

But $A^m B^n + B^m A^n = A^m B^n + A^n B^m = A^n B^m + A^m B^n$ (since $A$ and $B$ are real numbers).

Recall that $A B = 1$, so $B^m = (A B)^m / A^m = A^m / A^m = 1 / A^m$ only if $A B = 1$, but more generally, $B^m = (A B)^m / A^m = 1 / A^m$.

But let's use the following identity:

$A^m B^n + A^n B^m = (A B)^n A^{m-n} + (A B)^m B^{n-m}$

But since $A B = 1$, $(A B)^n = 1$, so $A^m B^n = A^{m-n}$, and $A^n B^m = B^{m-n}$.

Therefore,

$$a_m a_n = A^{m+n} + B^{m+n} + A^{m-n} + B^{m-n} = a_{m+n} + a_{m-n}$$

Thus, $a_{m+n} + a_{m-n} = a_m a_n$, as required.