25. Define $b_{n}=\left|a_{n+1}-a_{n}\right|$ for $n \geq 1$. From the equalities $a_{n+1}=b_{n-1}+b_{n-2}$, from $a_{n}=b_{n-2}+b_{n-3}$ we obtain $b_{n}=\left|b_{n-1}-b_{n-3}\right|$. From this relation we deduce that $b_{m} \leq \max \left(b_{n}, b_{n+1}, b_{n+2}\right)$ for all $m \geq n$, and consequently $b_{n}$ is bounded.
Lemma. If $\max \left(b_{n}, b_{n+1}, b_{n+2}\right)=M \geq 2$, then $\max \left(b_{n+6}, b_{n+7}, b_{n+8}\right) \leq$ $M-1$.
Proof. Assume the opposite. Suppose that $b_{j}=M, j \in\{n, n+1, n+2\}$, and let $b_{j+1}=x$ and $b_{j+2}=y$. Thus $b_{j+3}=M-y$. If $x, y, M-y$ are all less than $M$, then the contradiction is immediate. The remaining cases are these:
(i) $x=M$. Then the sequence has the form $M, M, y, M-y, y, \ldots$. and since $\max (y, M-y, y)=M$, we must have $y=0$ or $y=M$.
(ii) $y=M$. Then the sequence has the form $M, x, M, 0, x, M-x, \ldots$, and since $\max (0, x, M-x)=M$, we must have $x=0$ or $x=M$.
(iii) $y=0$. Then the sequence is $M, x, 0, M, M-x, M-x, x, \ldots$, and since $\max (M-x, x, x)=M$, we have $x=0$ or $x=M$.
In every case $M$ divides both $x$ and $y$. From the recurrence formula $M$ also divides $b_{i}$ for every $i<j$. However, $b_{2}=12^{12}-11^{11}$ and $b_{4}=11^{11}$ are relatively prime, a contradiction.
From $\max \left(b_{1}, b_{2}, b_{3}\right) \leq 13^{13}$ and the lemma we deduce inductively that $b_{m} \leq 1$ for all $n \geq 6-13^{13}-5$. Hence $a_{n}=b_{n-2}+b_{n-3}$ takes only the values $0,1,2$ for $n \geq 6 \cdot 13^{13}-2$. In particular, $a_{14}^{14}$ is 0 , 1 , or 2 . On the other hand, the sequence $a_{n}$ modulo 2 is as follows: $1,0,1,0,0,1,1 ; 1,0,1,0, \ldots$; and therefore it is periodic with period 7. Finally, $14^{14} \equiv 0$ modulo 7 , from which we obtain $\alpha_{14^{14}} \equiv a_{7} \equiv 1(\bmod 2)$. Therefore $\alpha_{14^{14}}=1$.
25. N3 (GBR) Let $a_{1}=11^{11}, a_{2}=12^{12}, a_{3}=13^{13}$, and
$$
a_{n}=\left|a_{n-1}-a_{n-2}\right|+\left|a_{n-2}-a_{n-3}\right|, \quad n \geq 4 \text {. }
$$

Determine $\alpha_{14^{14}}$.