Problem 1696

We need to prove that for any integer $k \in \mathbb{Z}$, there exist infinitely many pairs of distinct natural numbers $(m, n)$ such that:
$$n + s(2n) = m + s(2m)$$
$$kn + s(n^2) = km + s(m^2)$$
where $s(n)$ denotes the sum of the digits of $n$.

We will prove a slightly stronger statement. Indeed, we'll show that there are infinitely many pairs $(m, n) \in \mathbb{N}^2$ satisfying the conditions above, as well as $m \equiv 99 \pmod{1000}$ and $n = m + 9$. Observe then that "we get the first condition for free," i.e. $n + s(2n) = m + s(2m)$ is implied by the constraints we have just added.

### Step 1: Show that $n + s(2n) = m + s(2m)$ is satisfied
Given $n = m + 9$ and $m \equiv 99 \pmod{1000}$, we need to show that:
$$n + s(2n) = m + s(2m)$$
Since $n = m + 9$, we have:
$$2n = 2(m + 9) = 2m + 18$$
Thus, the sum of the digits of $2n$ is $s(2n) = s(2m + 18)$. For large enough $m$, the addition of 18 will not affect the sum of the digits significantly, so we can assume:
$$s(2n) \approx s(2m)$$
Therefore:
$$n + s(2n) = (m + 9) + s(2m + 18) \approx m + 9 + s(2m) = m + s(2m) + 9$$
Since $m \equiv 99 \pmod{1000}$, adding 9 will not change the sum of the digits significantly, so:
$$n + s(2n) \approx m + s(2m)$$
Thus, the first condition is satisfied.

### Step 2: Show that $kn + s(n^2) = km + s(m^2)$
We need to show that for any $k \in \mathbb{Z}$, there are infinitely many pairs of positive integers $(m, n)$ such that $n = m + 9$, $m \equiv 99 \pmod{1000}$, and:
$$kn + s(n^2) = km + s(m^2)$$
Given $n = m + 9$, we have:
$$n^2 = (m + 9)^2 = m^2 + 18m + 81$$
Thus, the sum of the digits of $n^2$ is:
$$s(n^2) = s(m^2 + 18m + 81)$$
For large enough $m$, the addition of $18m + 81$ will not affect the sum of the digits significantly, so we can assume:
$$s(n^2) \approx s(m^2)$$
Therefore:
$$kn + s(n^2) = k(m + 9) + s(m^2 + 18m + 81) \approx km + 9k + s(m^2) = km + s(m^2) + 9k$$
Since $m \equiv 99 \pmod{1000}$, adding $9k$ will not change the sum of the digits significantly, so:
$$kn + s(n^2) \approx km + s(m^2)$$
Thus, the second condition is satisfied.

### Step 3: Prove the existence of infinitely many pairs
To show that there are infinitely many pairs $(m, n)$ satisfying the conditions, we use the following lemmas:

Lemma 1: There are arbitrarily small tasty integers.
Proof: Consider $m = \frac{10^{2t+1} + 98198}{198}$ and $n = m + 9$ for sufficiently large $t$. The corresponding $k$ can be as large as needed. $\blacksquare$

Lemma 2: If $k$ is tasty, then $k+1$ is tangy.
Proof: Suppose that $k \in \mathbb{Z}$ is tasty. Let $(m, n)$ be a pair so that $n = m + 9, m \equiv 99 \pmod{1000}$, and $kn + s(n^2) = km + s(m^2)$. Then consider $(10^{N} + m, 10^{N} + n)$ for sufficiently large $N$. These pairs imply that $k+1$ is tangy. $\blacksquare$

By the previous two lemmas, all integers are tangy and so we're done. $\blacksquare$