Subfield: Number Theory


Question: Let $f$ be a function from the set of integers to the set of positive integers. Suppose that for any two integers $m$ and $n$, the difference $f(m)-f(n)$ is divisible by $f(m-n)$. Prove that for all integers $m, n$ with $f(m) \leq f(n)$ the number $f(n)$ is divisible by $f(m)$.

Solution:
Suppose that $x$ and $y$ are two integers with $f(x)<f(y)$. We will show that $f(x) \mid f(y)$. By taking $m=x$ and $n=y$ we see that

$$
f(x-y)|| f(x)-f(y) \mid=f(y)-f(x)>0
$$

so $f(x-y) \leq f(y)-f(x)<f(y)$. Hence the number $d=f(x)-f(x-y)$ satisfies

$$
-f(y)<-f(x-y)<d<f(x)<f(y) .
$$

Taking $m=x$ and $n=x-y$ we see that $f(y) \mid d$, so we deduce $d=0$, or in other words $f(x)=f(x-y)$. Taking $m=x$ and $n=y$ we see that $f(x)=f(x-y) \mid f(x)-f(y)$, which implies $f(x) \mid f(y)$.
We split the solution into a sequence of claims; in each claim, the letters $m$ and $n$ denote arbitrary integers.

Claim 1. $f(n) \mid f(m n)$.

Proof. Since trivially $f(n) \mid f(1 \cdot n)$ and $f(n) \mid f((k+1) n)-f(k n)$ for all integers $k$, this is easily seen by using induction on $m$ in both directions.

Claim 2. $f(n) \mid f(0)$ and $f(n)=f(-n)$.

Proof. The first part follows by plugging $m=0$ into Claim 1. Using Claim 1 twice with $m=-1$, we get $f(n)|f(-n)| f(n)$, from which the second part follows.

From Claim 1, we get $f(1) \mid f(n)$ for all integers $n$, so $f(1)$ is the minimal value attained by $f$. Next, from Claim 2, the function $f$ can attain only a finite number of values since all these values divide $f(0)$.

Now we prove the statement of the problem by induction on the number $N_{f}$ of values attained by $f$. In the base case $N_{f} \leq 2$, we either have $f(0) \neq f(1)$, in which case these two numbers are the only values attained by $f$ and the statement is clear, or we have $f(0)=f(1)$, in which case we have $f(1)|f(n)| f(0)$ for all integers $n$, so $f$ is constant and the statement is obvious again.

For the induction step, assume that $N_{f} \geq 3$, and let $a$ be the least positive integer with $f(a)>f(1)$. Note that such a number exists due to the symmetry of $f$ obtained in Claim 2 .



Claim 3. $f(n) \neq f(1)$ if and only if $a \mid n$.

Proof. Since $f(1)=\cdots=f(a-1)<f(a)$, the claim follows from the fact that

$$
f(n)=f(1) \Longleftrightarrow f(n+a)=f(1) .
$$

So it suffices to prove this fact.

Assume that $f(n)=f(1)$. Then $f(n+a) \mid f(a)-f(-n)=f(a)-f(n)>0$, so $f(n+a) \leq$ $f(a)-f(n)<f(a)$; in particular the difference $f(n+a)-f(n)$ is stricly smaller than $f(a)$. Furthermore, this difference is divisible by $f(a)$ and nonnegative since $f(n)=f(1)$ is the least value attained by $f$. So we have $f(n+a)-f(n)=0$, as desired. For the converse direction we only need to remark that $f(n+a)=f(1)$ entails $f(-n-a)=f(1)$, and hence $f(n)=f(-n)=f(1)$ by the forward implication.

We return to the induction step. So let us take two arbitrary integers $m$ and $n$ with $f(m) \leq f(n)$. If $a \nmid m$, then we have $f(m)=f(1) \mid f(n)$. On the other hand, suppose that $a \mid m$; then by Claim $3 a \mid n$ as well. Now define the function $g(x)=f(a x)$. Clearly, $g$ satisfies the conditions of the problem, but $N_{g}<N_{f}-1$, since $g$ does not attain $f(1)$. Hence, by the induction hypothesis, $f(m)=g(m / a) \mid g(n / a)=f(n)$, as desired.