8. (5th "Hope Cup" Competition Question) In the sequence $\left\{a_{n}\right\}$, $a_{1}=13, a_{2}=56$, for all positive integers $n$ there is $a_{n+1}=a_{n}+a_{n+2}$, then $a_{1934}=$ $\qquad$ .
8. (Iterative Elimination Method)

Since $a_{n+1}=a_{n}+a_{n+2}$,
(1)
then $a_{n+2}=a_{n-1}+a_{n+3}$.
(1) + (2) gives
$$
a_{n+3}=-a_{n} \text {. }
$$

So $a_{n+6}=-a_{n+3}=a_{n}$.
Therefore, 6 is the period of the sequence $\left\{a_{n}\right\}$. Thus
$$
a_{1391}=a_{6 \times 231-2}=a_{2}=56 .
$$