Problem (Number Theory):
For every positive integer $n$ with prime factorization $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, define

$$
\mho(n)=\sum_{i: p_{i}>10^{100}} \alpha_{i}\tag{1}
$$

That is, $\mho(n)$ is the number of prime factors of $n$ greater than $10^{100}$, counted with multiplicity.

Find all strictly increasing functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that

$$
\mho(f(a)-f(b)) \leqslant \mho(a-b) \quad \text { for all integers } a \text { and } b \text { with } a>b \text {. }
$$

Solution:
A straightforward check shows that all the functions listed in the answer satisfy the problem condition. It remains to show the converse.

Assume that $f$ is a function satisfying the problem condition. Notice that the function $g(x)=f(x)-f(0)$ also satisfies this condition. Replacing $f$ by $g$, we assume from now on that $f(0)=0$; then $f(n)>0$ for any positive integer $n$. Thus, we aim to prove that there exists a positive integer $a$ with $\mho(a)=0$ such that $f(n)=a n$ for all $n \in \mathbb{Z}$.

We start by introducing some notation. Set $N=10^{100}$. We say that a prime $p$ is large if $p>N$, and $p$ is small otherwise; let $\mathcal{S}$ be the set of all small primes. Next, we say that a positive integer is large or small if all its prime factors are such (thus, the number 1 is the unique number which is both large and small). For a positive integer $k$, we denote the greatest large divisor of $k$ and the greatest small divisor of $k$ by $L(k)$ and $S(k)$, respectively; thus, $k=L(k) S(k)$.

We split the proof into three steps.

Step 1. We prove that for every large $k$, we have $k|f(a)-f(b) \Longleftrightarrow k| a-b$. In other $\overline{\text { words, }} L(f(a)-f(b))=L(a-b)$ for all integers $a$ and $b$ with $a>b$.

We use induction on $k$. The base case $k=1$ is trivial. For the induction step, assume that $k_{0}$ is a large number, and that the statement holds for all large numbers $k$ with $k<k_{0}$.

Claim 1. For any integers $x$ and $y$ with $0<x-y<k_{0}$, the number $k_{0}$ does not divide $f(x)-f(y)$.

Proof. Assume, to the contrary, that $k_{0} \mid f(x)-f(y)$. Let $\ell=L(x-y)$; then $\ell \leqslant x-y<k_{0}$. By the induction hypothesis, $\ell \mid f(x)-f(y)$, and thus $\operatorname{lcm}\left(k_{0}, \ell\right) \mid f(x)-f(y)$. Notice that $\operatorname{lcm}\left(k_{0}, \ell\right)$ is large, and $\operatorname{lcm}\left(k_{0}, \ell\right) \geqslant k_{0}>\ell$. But then

$$
\mho(f(x)-f(y)) \geqslant \mho\left(\operatorname{lcm}\left(k_{0}, \ell\right)\right)>\mho(\ell)=\mho(x-y),
$$

which is impossible.

Now we complete the induction step. By Claim 1, for every integer $a$ each of the sequences

$$
f(a), f(a+1), \ldots, f\left(a+k_{0}-1\right) \quad \text { and } \quad f(a+1), f(a+2), \ldots, f\left(a+k_{0}\right)
$$

forms a complete residue system modulo $k_{0}$. This yields $f(a) \equiv f\left(a+k_{0}\right)\left(\bmod k_{0}\right)$. Thus, $f(a) \equiv f(b)\left(\bmod k_{0}\right)$ whenever $a \equiv b\left(\bmod k_{0}\right)$.

Finally, if $a \not \equiv b\left(\bmod k_{0}\right)$ then there exists an integer $b^{\prime}$ such that $b^{\prime} \equiv b\left(\bmod k_{0}\right)$ and $\left|a-b^{\prime}\right|<k_{0}$. Then $f(b) \equiv f\left(b^{\prime}\right) \not \equiv f(a)\left(\bmod k_{0}\right)$. The induction step is proved.

Step 2. We prove that for some small integer a there exist infinitely many integers $n$ such that $\overline{f(n)}=$ an. In other words, $f$ is linear on some infinite set.

We start with the following general statement.



Claim 2. There exists a constant $c$ such that $f(t)<c t$ for every positive integer $t>N$.

Proof. Let $d$ be the product of all small primes, and let $\alpha$ be a positive integer such that $2^{\alpha}>f(N)$. Then, for every $p \in \mathcal{S}$ the numbers $f(0), f(1), \ldots, f(N)$ are distinct modulo $p^{\alpha}$. Set $P=d^{\alpha}$ and $c=P+f(N)$.

Choose any integer $t>N$. Due to the choice of $\alpha$, for every $p \in \mathcal{S}$ there exists at most one nonnegative integer $i \leqslant N$ with $p^{\alpha} \mid f(t)-f(i)$. Since $|\mathcal{S}|<N$, we can choose a nonnegative integer $j \leqslant N$ such that $p^{\alpha} \nmid f(t)-f(j)$ for all $p \in \mathcal{S}$. Therefore, $S(f(t)-f(j))<P$.

On the other hand, Step 1 shows that $L(f(t)-f(j))=L(t-j) \leqslant t-j$. Since $0 \leqslant j \leqslant N$, this yields

$$
f(t)=f(j)+L(f(t)-f(j)) \cdot S(f(t)-f(j))<f(N)+(t-j) P \leqslant(P+f(N)) t=c t
$$

Now let $\mathcal{T}$ be the set of large primes. For every $t \in \mathcal{T}$, Step 1 implies $L(f(t))=t$, so the ratio $f(t) / t$ is an integer. Now Claim 2 leaves us with only finitely many choices for this ratio, which means that there exists an infinite subset $\mathcal{T}^{\prime} \subseteq \mathcal{T}$ and a positive integer $a$ such that $f(t)=a t$ for all $t \in \mathcal{T}^{\prime}$, as required.

Since $L(t)=L(f(t))=L(a) L(t)$ for all $t \in \mathcal{T}^{\prime}$, we get $L(a)=1$, so the number $a$ is small.

Step 3. We show that $f(x)=$ ax for all $x \in \mathbb{Z}$.

Let $R_{i}=\{x \in \mathbb{Z}: x \equiv i(\bmod N !)\}$ denote the residue class of $i$ modulo $N !$.

Claim 3. Assume that for some $r$, there are infinitely many $n \in R_{r}$ such that $f(n)=a n$. Then $f(x)=a x$ for all $x \in R_{r+1}$.

Proof. Choose any $x \in R_{r+1}$. By our assumption, we can select $n \in R_{r}$ such that $f(n)=$ an and $|n-x|>|f(x)-a x|$. Since $n-x \equiv r-(r+1)=-1(\bmod N !)$, the number $|n-x|$ is large. Therefore, by Step 1 we have $f(x) \equiv f(n)=a n \equiv a x(\bmod n-x)$, so $n-x \mid f(x)-a x$. Due to the choice of $n$, this yields $f(x)=a x$.

To complete Step 3, notice that the set $\mathcal{T}^{\prime}$ found in Step 2 contains infinitely many elements of some residue class $R_{i}$. Applying Claim 3, we successively obtain that $f(x)=a x$ for all $x \in R_{i+1}, R_{i+2}, \ldots, R_{i+N !}=R_{i}$. This finishes the solution.

Final answer: $f(x)=a x+b$, where $b$ is an arbitrary integer, and $a$ is an arbitrary positive integer with $\mho(a)=0$