Question 85: Find all pairs of positive integers $(m, a)$ such that the sequence $\left\{a_{n}\right\}_{n \in Z^{+}}: a_{1}=a, a_{n+1}=$ $\left\{\begin{array}{c}a_{n}^{2}+2^{m}, \text { when } a_{n}<2^{m} \\ \frac{a_{n}}{2}, \text { when } a_{n} \geq 2^{m}\end{array}\right.$ consists of integers for all terms.
The 85th question:
Solution: The required pairs of positive integers $(\mathrm{m}, \mathrm{a})=\left(2,2^{\mathrm{k}}\right), \mathrm{k} \in \mathrm{Z}^{+}$.
For any positive integer $n$, let $c_{n}=v_{2}\left(a_{n}\right), b_{n}=\frac{a_{n}}{\left.2^{2} a_{n}\right)}$, then $a_{n}=b_{n} \cdot 2^{c_{n}}$. We first prove two lemmas.
Lemma 1: For any positive integer $\mathrm{n}$, we have $\mathrm{b}_{\mathrm{n}} \leq 2^{\mathrm{m}}$.
Proof of Lemma 1: If there exists $b_{n}>2^{m}$, let the subsequence of terms in $\left\{b_{n}\right\}$ that are greater than $2^{m}$ be $b_{n_{1}}, b_{n_{2}}, \ldots$. Consider the smallest term in the sequence $\left\{c_{n_{i}}\right\}$, say $c_{n_{j}}$. Since $a_{n_{j}} \geq b_{n_{j}}>2^{m}$, according to the condition, $a_{n_{j}+1}=\frac{a_{n_{j}}}{2}=b_{n_{j}} \cdot 2^{c_{n_{j}}-1}$, which implies $b_{n_{j}+1}=b_{n_{j}}>2^{m}$, and $c_{n_{j}+1}=c_{n_{j}}-1$. Then $a_{n+1}=\left(b_{n}^{2} \cdot 2^{2 c_{n}-m}+1\right) 2^{m}$, so $b_{n+1}=b_{n}^{2} \cdot 2^{2 c_{n}-m}+1>b_{n}$;
(2) If $2 c_{n}b_{n}$;
(3) If $2 c_{n}=m$, then $a_{n+1}=\frac{b_{n}^{2}+1}{2} \cdot 2^{m+1}$, so $b_{n+1}=\frac{b_{n}^{2}+1}{2} \geq b_{n}$, equality holds if and only if $b_{n}=$
1.
In summary, Lemma 2 is proved.
Now, let's return to the original problem.
According to Lemmas 1 and 2, the sequence $\left\{b_{n}\right\}$ must eventually become a constant sequence. That is, there exists a positive integer $N$, such that for $n \geq N$, we have $\mathrm{b}_{\mathrm{n}}=\mathrm{b}$ as a constant.

If $a_{n} \geq 2^{m}$, then $a_{n+1}$ reduces to $\frac{a_{n}}{2}$, so the sequence $\left\{a_{n}\right\}$ contains infinitely many terms less than 2 m. Take any positive integer $i>N$ such that $a_{i}<2^{m}$, then $b_{i+1}=b_{i}=b$, according to the discussion in Lemma 2, it can only be that $b=1$, and $2 c_{i}=m \Rightarrow c_{i}=\frac{m}{2}$, thus $\mathrm{a}_{\mathrm{i}}=2^{\frac{\mathrm{m}}{2}}$. Furthermore, according to Lemma 2, $\left\{\mathrm{b}_{\mathrm{n}}\right\}$ is non-decreasing, so for any positive integer $\mathrm{n}$, we have $\mathrm{b}_{\mathrm{n}}=1$, that is, every term in the sequence $\left\{\mathrm{a}_{\mathrm{n}}\right\}$ is a power of 2. $2^{m}$ is a power of 2, thus $2 m-2=m \Rightarrow m=2$.
From the above analysis, $\mathrm{a}_{1}=\mathrm{a}$ is also a power of 2, let $\mathrm{a}=2^{\mathrm{k}}$. When $k \geq 1$, the sequence $\left\{\mathrm{a}_{\mathrm{n}}\right\}$ eventually becomes:
$2, 8, 4$ in a cycle; when $k=0$, the sequence $\left\{\mathrm{a}_{\mathrm{n}}\right\}$ is: $1, 5, \frac{5}{2}$, which does not satisfy the condition. In summary, the required $(\mathrm{m}, \mathrm{a})=\left(2,2^{\mathrm{k}}\right), \mathrm{k} \in \mathrm{Z}^{+}$.