Problem 1617

1. **Prove that $5$ is a nice number.**

Given: $kn + m \mid km + n$ for $k = 5$, we need to show that $n \mid m$.

Let's start by substituting $k = 5$:
$$5n + m \mid 5m + n$$

Suppose $\gcd(m, n) = d$, then we can write $m = da$ and $n = db$ where $\gcd(a, b) = 1$. Substituting these into the divisibility condition, we get:
$$5(db) + da \mid 5(da) + db$$
Simplifying, we have:
$$d(5b + a) \mid d(5a + b)$$
Since $d$ is a common factor, we can divide both sides by $d$:
$$5b + a \mid 5a + b$$

Now, we need to show that $5b + a \mid 5a + b$. Consider the expression:
$$5a + b = (5b + a) \cdot q + r$$
where $q$ is the quotient and $r$ is the remainder. Since $5b + a \mid 5a + b$, the remainder $r$ must be zero. Therefore:
$$5a + b = (5b + a) \cdot q$$

Rearranging, we get:
$$5a + b - (5b + a) \cdot q = 0$$
Simplifying further:
$$5a + b - 5bq - aq = 0$$
$$5a + b = 5bq + aq$$
$$5a + b = q(5b + a)$$

Since $\gcd(a, b) = 1$, the only way this can hold is if $q = 1$ and $5b + a = 5a + b$. This implies:
$$5b + a = 5a + b$$
$$4b = 4a$$
$$b = a$$

Therefore, $n = db$ and $m = da$ imply $n \mid m$.

2. Find all the nice numbers.

We need to find all $k$ such that for any pair $(m, n)$ of positive integers satisfying $kn + m \mid km + n$, we have $n \mid m$.

From the previous part, we know that $k = 5$ is a nice number. Let's generalize this for any $k$.

Given $kn + m \mid km + n$, we need to show that $n \mid m$.

Suppose $\gcd(m, n) = d$, then $m = da$ and $n = db$ where $\gcd(a, b) = 1$. Substituting these into the divisibility condition, we get:
$$k(db) + da \mid k(da) + db$$
Simplifying, we have:
$$d(kb + a) \mid d(ka + b)$$
Since $d$ is a common factor, we can divide both sides by $d$:
$$kb + a \mid ka + b$$

Now, we need to show that $kb + a \mid ka + b$. Consider the expression:
$$ka + b = (kb + a) \cdot q + r$$
where $q$ is the quotient and $r$ is the remainder. Since $kb + a \mid ka + b$, the remainder $r$ must be zero. Therefore:
$$ka + b = (kb + a) \cdot q$$

Rearranging, we get:
$$ka + b - (kb + a) \cdot q = 0$$
Simplifying further:
$$ka + b - kbq - aq = 0$$
$$ka + b = q(kb + a)$$

Since $\gcd(a, b) = 1$, the only way this can hold is if $q = 1$ and $kb + a = ka + b$. This implies:
$$kb + a = ka + b$$
$$(k-1)b = (k-1)a$$
$$b = a$$

Therefore, $n = db$ and $m = da$ imply $n \mid m$.

To find all nice numbers, we need to check the condition $k > 2$. If $k > 2$, then we choose $a = 2k - 1$ and $b = k - 2$:
$$kb + a = k(k-2) + (2k-1) = k^2 - 2k + 2k - 1 = k^2 - 1$$
$$k^2 - 1 \mid k^2 - 1$$
$$k - 2 \mid 2k - 1$$
$$k - 2 \mid 3$$
$$k \in \{3, 5\}$$

Therefore, the nice numbers are $2, 3, 5$.

The final answer is $\boxed{ 2, 3, 5 }$.