Problem 1773

For any $h = 2^{r}$ (where $r$ is a non-negative integer), we need to find all $k \in \mathbb{N}$ which satisfy the following condition: There exists an odd natural number $m > 1$ and $n \in \mathbb{N}$, such that $k \mid m^{h} - 1$ and $m \mid n^{\frac{m^{h}-1}{k}} + 1$.

We claim that $k$ works if and only if $2^{r+1} \mid k$.

### Necessity:
Let $A = \frac{m^{2^r} - 1}{k}$. We need to show that $\nu_2(p-1) \geq \nu_2(A) + 1$ for any prime divisor $p$ of $m$.

Proof: Let $u = \mathrm{ord}_p(n)$. Then $u \mid 2A$ but $u \nmid A$ (since $m$ is odd). Therefore, $\nu_2(u) = \nu_2(A) + 1$. Since $u \mid p-1$, we have $\nu_2(p-1) \geq \nu_2(A) + 1$.

Let $t = \nu_2(A)$. The claim implies $m \equiv 1 \pmod{2^{t+1}}$. Using the Lifting The Exponent (LTE) lemma, we get:
$$t = \nu_2(m^{2^r} - 1) - \nu_2(k) = \nu_2(m-1) + \nu_2(m+1) + r - 1 - \nu_2(k).$$
Since $\nu_2(m-1) \geq t + 1$ and $\nu_2(m+1) \geq 1$, we have:
$$t \geq (t + 1) + 1 + r - 1 - \nu_2(k),$$
which simplifies to $\nu_2(k) \geq r + 1$.

### Sufficiency:
By Dirichlet's theorem, take a prime $p \equiv 1 \pmod{4k}$ and let $m = p$. The first divisibility condition $k \mid p^{2^r} - 1$ is clearly satisfied. Let $t = \nu_2\left(\frac{p^{2^r} - 1}{k}\right)$. Since $p + 1 \equiv 2 \pmod{4}$, we have:
$$t = \nu_2(p-1) + \nu_2(p+1) + r - 1 - \nu_2(k) = \nu_2(p-1) + (r - \nu_2(k)).$$
Thus, $t + 1 \leq \nu_2(p-1)$. Since the groups $\mathbb{Z}_p^{\times}$ and $\mathbb{Z}_{p-1}$ are isomorphic, there exists an element $n$ of $\mathbb{Z}_p^{\times}$ that has order $2^{t+1}$. This means:
$$n^{2^t} \equiv -1 \pmod{p} \implies n^{\frac{p^{2^r} - 1}{k}} = n^{2^t \cdot \text{odd}} \equiv -1 \pmod{p},$$
so this $m$ and $n$ work.

The answer is: \boxed{2^{r+1}}.