-- $f(n)|f(0)$ for every integer $n$ $f(m)=f(m-0)|f(m)-f(0)$. Then $f(m)|f(0)$ -- $f(-n) = f(n)$ for every integer $n$ $f(-n)=f(0-n)|f(0)-f(n)$. Then $f(-n)|f(n)$ since $f(-n)|f(0)$. Similarly $f(n)|f(-n)$. Then $f(n)=f(-n)$ Suppose that $f(m) < f(n)$ and $f(m)$ doesn't divide $f(n)$. $f(m+n)=f(m-(-n))|f(m)-f(-n)=f(m)-f(n)$. Then $f(m+n)\le f(n)-f(m) \ \ \ \ (*)$ $f(m)|f(n+m)-f(n)$. Then $f(m)$ doesn't divide $f(n+m)$. Then $|f(n+m)-fm)|$ is not $0$ $f(n)|f(n+m)-f(m)$. Then $f(n)\le f(n+m)-f(m)$ or $f(n)\le f(m)-f(n+m)$ If $f(n)\le f(n+m)-f(m)$, then $f(n)+f(m+n)\le f(m+n)+f(n)-2f(m)$ (adding with (*)), which is impossible If $f(n)\le f(m)-f(n+m)$, then $f(n)+f(n+m)\le f(n)-f(n+m)$ (adding with (*)), which is impossible Therefore if $f(m)\le f(n)$, we must have that $f(m) |f(n)$

here is the sketch. 1. f(x) l f(0) 2. f(x) l f(-x), f(-x) l f(x) 3. f(x)=f(-x) 4.f(x+y) l f(x)+f(y), f(x) l f(x+y)-f(y), f(y) l f(x+y)-f(x) 5. two of .f(x+y), f(x), f(y) are the same. 6. f(a) l f(b) or f(b) l f(a) if a,b are x, y or x+y

Sorry but,is it true? f(m)|f(m n)-f(n),then f(m)|f(2m).,and f(2m)=f(m m)|f(m)-f(-m),so f(m)|f(-m) for all integer m so f(m)= f(-m) for all m. We have f(m n)|f(m)-f(-n)=f(m)-f(n) for all m,n >0. Then |f(m)-f(n)|>=f(m n) (1). IF f(m n)>f(n) for m,n>0 then sine f(m)|f(m n)-f(n) we have f(m n)-f(n)>=f(m) (2) Sine (1) and (2) :f(m) f(n)<=|f(m)-f(n)|Contracdiction!So f(m)>=f(n) for all m<=n. Pleas

Solution: We have $f(m-n)|f(m)-f(n)$ setting $n=0$ gives $f(m)|f(0)$, setting $m=0$ we get $f(-n)|f(n)$ implying $f(n)|f(-n)$ hence $f(-n)= f(n)$ .This gives $f(m+n)|f(m)-f(n)$ , If $f(n)\geq f(m)$ we have $f(m+n)\leq f(n)-f(m)$.Further $f(n)|f(m+n)-f(m)$.This implies we may have $(i) f(n)\leq f(m+n)-f(m)$ but from the previous inequality $f(m+n)+f(m)\leq f(n)\leq f(m+n)-f(m)$ which means $2f(m)\leq 0$ clearly impossible.We consider case $(ii) f(n)\leq f(m)-f(m+n)$, but this ,with $f(n)\geq f(m)$, gives $f(m)+f(m+n)\leq f(n)+f(m+n)\leq f(m)$,giving $f(m+n)\leq 0$, again not possible.The last case $(iii)$ and the only possibility is $f(m+n)= f(m)$.But we have $f(m)|f(m+n)-f(n)$implying $f(m)|f(m)-f(n)$. Hence we must have $f(m)|f(n)$, and we are done!

Here's my convoluted solution I'm not totally sure about. If anyone takes the time and checks this stream of consciousness, I'll be happy. First, $n=0$ -> $f(m)|f(0)$ for every integer $m$. Then, since we have $f(1)|f(1), f(1)|f(2)-f(1), f(1)|f(3)-f(2), ... $ for both positive and negative integers, for every integer $m$ we have $f(1)|f(m)$. Therefore we can replace our function with $f_1(x)=f(x)/f(1)$, making sure this affects neither the conditions nor the required result. Similarly, we get $f(-1)=1$. Next, using the basic statement we have that for all integral variables $f(m)|f(km)$. Since the function is positive, employing $k=-1$ we obtain it's even. Now suppose there's number $m$ for which $f(m)>1$ and $f(m-1)>1$. Substituting $1$ in the place of $n$, we have that $f(m-1)|f(m)-1$, and doing the same with $m-1$ and $-1$, $f(m)|f(m-1)-1$. Hence, at least one value, $f(m-1), f(m)$ must equal 1 and we have a contradiction. Denote the set of all integers $m$ for which $f(m)>1$ as $P$. As shown above, $m \in P$ entails $km \in P$. Now we'll show that $m \in P$ and $n \in P$ entail $(m+n) \in P$. Really, if $f(m+n)=1$, then according to the statement $f(n)|f(m)-1$ and $f(m)|f(n)-1$, hence, either of these values is equal to one, which is against our assumption. So, if $f(m)>1, f(n)>1, f(p)>1$, for any integer multipliers $f(am+bn+cp)>1$ (*). On the other hand, no pair of coprime numbers can be contained in $P$ - otherwise we can find two neighbouring numbers each of them dividing one of the taken. Given all this, we conclude that all numbers of $P$ share a divisor $s$ greater than $1$. [If a set of numbers has no common divisor, there exists a number that is coprime with all of them and can be expressed as their combination like (*)]. In this case, we introduce a new function $f_2(x)=f(x/s)$ and continue the same steps for it. Numbers which are not divisible by $s$ can be disregarded, because they satisfy the needed condition anyway. Step by step, every number will be reached in this way at some point and our construction proves what is necessary.

Surely your example is not good, since $f$ takes positive values. However, the post above you opens some hope on such a question. Definitely $f$ is even, the fibres of $f$ are finitely many, and $\textrm{Im} f$ is a finite chain $C$ in $(\mathbb{Z}_+^*, \mid)$, with $\min C = f(\pm 1)$ and $\max C = f(0)$. How much definition can we bring as to the structure of $f$?

I believe the functions can be characterized as follows: Let $1 = a_1 < a_2 < \cdots < a_t$ be integers, with $a_1 \, | \, a_2 \, | \, \cdots \, | \, a_t$, and let $c_1 < c_2 < \ldots < c_t$ be any integers such that $c_1 \, | \, c_2 \, | \, \cdots \, | \, c_t$. For any nonzero integer $k$, let $\alpha(k)$ be the largest integer $i$ such that $a_i | k$. Then $f(k) =c_{\alpha(k)}$ for all nonzero $k$, and $f(0)$ is any multiple of $c_t$. Please inform me if the following solution has any errors:

a different approach: 1. $f(n)|f(0)$ 2. $f(n)=f(-n)$ 3. $f(n)|f(kn)$ 4. let $am+bn=gcd(m,n)$ then $f(bn)|f(am)-f(gcd(m,n))$ $f(am)|f(bn)-f(gcd(m,n))$ at least one of $f(am)-f(gcd(m,n))$ , $f(bn)-f(gcd(m,n))$ equals 0 we have $f(am) \ge f(m) \ge f(gcd(m,n))$ $f(n) \ge f(m)$ so $f(am)=f(m)=f(gcd(m,n))$ and $f(gcd(m,n))|f(n)$ we have $f(m)|f(n)$

orl wrote: 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 $m$ and $n$ with $f(m) \leq f(n)$, the number $f(n)$ is divisible by $f(m).$ Proposed by Mahyar Sefidgaran, Iran Nice problem. We use the facts: $a,b\in\mathbb{N}a|b\implies b\ge a$ and $a|b\pm a\implies a|b$. First let's prove that $f$ is an even function. Set $n=0$ giving that $f(m)|f(m)-f(0)\implies f(m)|f(0)\forall m$. Setting $m=0$, we get $f(-n)|f(n)$ and since $f$ is defined for all integers, it is symmetric for $n$ and $-n$ implying $f(-n)=f(n)$ Then expand using $m+n=m-(-n)$ to note that $f(m+n)|f(m)-f(n)$. Since the co-domain of $f$ is $\mathbb{N}$ we have three cases. Case 1: $f(m)-f(n)\ge f(m+n)$. But we have $f(m)=f(m+n-n)|f(m+n)-f(n)$ giving $f(m)-f(n)\ge f(m+n)\ge f(m)+f(n)$ $\implies f(m)+f(n)\le f(m)-f(n)\implies 2f(n)\le 0$, contradiction since $f(m)\in\mathbb{N}$. Case 2: $ f(m)\le f(n)-f(m+n) $ which is almost similar using the given fact $f(n)\ge f(m)$ and not possible. Case 3: $f(m+n)=f(m)$ holds as the only possibility. Then $f(m)|f(m+n)-f(m)=f(n)-f(m)\implies f(m)|f(n)$. We are done. Edit: Since most solutions use the same idea, I was thinking to solve this using induction. Is it possible with induction? Also, may be in the cases I have reversed the order of $m$ and $n$.

What do you think about this solution? 1. f(m) divides f(0) for all m integers. So, f(m) can take a finite number of values. 2. We can partition the integers into a finite number of sets S1, S2, ..., Sr so that for all i all elements of Si have same image f(Si) under f. 3. WLOG assume that f(S1)<f(S2)<...<f(Sr) 4. Since f(m-n)|f(m)-f(n), we can find 1<=t<=r so that f(St)|f(Sj)-f(Si) for all 1<=i<j<=r 5. We prove by induction on r that f(S1)|f(S2)|.....|f(Sr).

mszew wrote: "Is it possible to find all f?For example: $f(n) \in \{a,-a\}$ and $f(0)=k a$ Another example:$f(odd)=a$, $f(even)=ab$, $f(0)=abc$.

mathmdmb wrote: orl wrote: 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 $m$ and $n$ with $f(m) \leq f(n)$, the number $f(n)$ is divisible by $f(m).$ Proposed by Mahyar Sefidgaran, Iran Case 1: $f(m)-f(n)\ge f(m+n)$. But we have $f(m)=f(m+n-n)|f(m+n)-f(n)$ giving $f(m)-f(n)\ge f(m+n)\ge f(m)+f(n)$ $\implies f(m)+f(n)\le f(m)-f(n)\implies 2f(n)\le 0$, contradiction since $f(m)\in\mathbb{N}$. . Wrong f(m)|f(m+n)-f(n)=0 gives f(m)+f(n ge f(m+n) here with f; qnd f(m+n) different Hence it is wrong qnd zouldn`t get more than four and you hqve typo in casa2

Again, a fat-less late proof. Let us first gather some trivial results. Since $f(m)=f(m-0)\mid f(m)-f(0)$, it follows $f(m)\mid f(0)$ for every integer $m$. We now have $f(-n)=f(0-n)\mid f(0)-f(n)$, so $f(-n)\mid f(n)$, since $f(-n)\mid f(0)$. Then also $f(n)\mid f(-n)$, therefore $f(n)=f(-n)$ (they are both positive). Now we can write \[f(m-n) \mid f(m) - f(n),\] \[f(n) \mid f(m) - f(m-n),\] \[f(m) \mid f(n) - f(n-m) = f(n) - f(m-n),\] therefore we have three positive integers, each dividing the difference of the other two. That is only possible if two of them are equal, and dividing the third, so for $f(m) \leq f(n)$ we either have $f(m) = f(n)$, or else we can say even more, namely $f(m-n) = f(m) \mid f(n)$. What can we say about such functions? It is now time for some elementary group theory. Define the fibre $\Omega =f^{-1} (\omega)$ that contains $0$. In general, we must have $\Omega - \Omega \subseteq \Omega$, since for any elements $m,n \in \Omega$ we have $\omega = f(n) \mid f(m) - f(m-n) = \omega - f(m-n)$, implying $f(m-n) = \omega$, hence $m-n \in \Omega$. Therefore $\Omega$ is some additive subgroup $(k\mathbb{Z},+) \subsetneq (\mathbb{Z},+)$. Clearly, a constant $f$ fulfills. If $f$ takes more than one value, and it may only take finitely many, all divisors of $\omega = f(0)$ (so $\textrm{Im} f$ is a finite chain in the poset $(\mathbb{N}^*, \mid)$, of maximal element $\omega$), then it is easy to show that $\alpha = f(\pm 1)$ divides them all. A non-constant example with only two fibres $0\in \Omega =f^{-1} (\omega)$, $\pm 1 \in A=f^{-1} (\alpha)$ (with $\alpha \mid \omega$) will have to have $A = \mathbb{Z} \setminus \Omega$, with $\Omega$ some additive subgroup $(k\mathbb{Z},+) \subsetneq (\mathbb{Z},+)$. This can most likely be pursued even farther - I will append any further progress ...

wow, seems anyone who posts answers here get a six star. Here's my solution property 0$f(m-n)|f(m)-f(n)$ property 1$f(n)|f(0)$ for all $n$ property 2$f(-n)=f(n)$ for all $n$ property 3$f(1)|f(n)$ for all $n$ which implies $f(1)|f(n)|f(0)$ therefore $f(1) \le f(n) \le f(0)$ property 4 for all fixed $k$ $f(k)|f(kn)$ for all $n$, which implies $f(k) \le f(kn)$ property 5 $f(m+n)|f(m)-f(n)$ Use property 5, $f(3)|f(2)-f(1)$ then $Af(3)+f(1)=f(2)$ if $A \ge 1$ $f(3) \le f(2)-f(1)$ if $A \le 1$ $f(3) \ge f(2)-f(1)$, contradiction so $A \ge 1$ therefore $f(3) \le f(2)-f(1) \le f(2)$ $f(4)|f(3)-f(1)$ $f(4) \le f(3)-f(1) \le f(3) \le f(2)$ $f(5)|f(4)-f(1)$ $f(5) \le f(4)-f(1) \le f(4) \le f(3) \le f(2)$ But this contradicts property 4 so $f(m)=f(n)$ for all $m, n > 1$

sexynsmartjenny wrote: Use property 5, $f(3)|f(2)-f(1)$ then $Af(3)+f(1)=f(2)$ if $A \ge 1$ $f(3) \le f(2)-f(1)$ if $A \le 1$ $f(3) \ge f(2)-f(1)$, contradiction ... There is no contradiction, in general, when you do $f(m+1) \mid f(m) - f(1)$; if $f(1)=f(m)$ then you can draw no inference on $f(m+1)$ (except that $f(m) = f(1) \mid f(m+1)$). If $f(1) < f(m)$, then indeed it follows $f(m+1) \leq f(m) - f(1)$, in fact the stronger $f(m+1) = f(1) < f(m)$. So you get less stars, not so much for having botched your proof, but for not giving credit to those that gave examples other than the trivial function you claim is the solution, nor to the Problem Selection Committee, that should have discovered that just the trivial solution existed (and so must have disqualified it)

The definitive characterization of such functions follows (and is a direct consequence of my late considerations). Let $C = (1=k_1,k_2,\ldots,k_n = k)$ be a finite ascending chain in $(\mathbb{N}, \mid)$, with $n \geq 1$ and possibly $k=0$. Let $V = (\alpha = v_1,v_2,\ldots,v_n = \omega)$ be an associated finite ascending chain in $(\mathbb{N}^*, \mid)$ (thus with $\omega \neq 0$). For $n>1$ take $A_n = \Omega = k\mathbb{Z}$, $A_m = k_m\mathbb{Z} \setminus k_{m+1}\mathbb{Z}$, for $1\leq m \leq n-1$ (thus $A_1 = A = \mathbb{Z} \setminus k_{2}\mathbb{Z}$), partitioning $\mathbb{Z}$. Define $f(A_m) = v_m$, for $1\leq m \leq n$ (of course, for $n=1$ we have $f(\mathbb{Z}) = f(\Omega) = \omega$).

$f(m-n) \mid f(m)-f(n)$......(p) setting $n$ as $0$ in (p) we get $f(m) \mid f(0)$ setting $m$ as $0$ in (p) we get $f(-n) \mid f(n)$......(1) setting $m$ as $0$,$n$ as $-n$ in (p) we get $f(n) \mid f(-n)$......(2) so from (1) and (2) we can say $f(n)=f(-n)$......(3) setting $m$ as $2n$ in (p) we get $f(n) \mid f(2n)$ or,$f(2n) \geq f(n)$......(y) setting $n$ as $-n$ in (p) we get $f(m+n) \mid f(m)-f(n)$ or, $f(m)-f(n) \geq f(m+n)$......(x) setting $m$ as $(m+n)$ in (p) we get $f(m) \mid f(m+n)- f(n)$ if $f(m) \mid f(m+n)$ the proof is complete.so this will be our requirement from now on. there can be $3$ relations between $f(m+n)$ and $f(m)$ a.$f(m+n)> f(m)$ or,$f(m+n)>f(m)-f(n)$ but this is a contradiction according to (x) b.$f(m+n)<f(m)$ setting $m$ as $n$ yields $f(2n)<f(n)$ which is a contradiction according to (y) so $f(m+n)=f(m)$ hence $f(m) \mid f(m+n)$ which was our requirement.

SCP, I said in edit that it was due to my reversing their order, which was illegal indeed. Whatever, here is the clearer solution. $f(m-0)|f(m)-f(0)\implies f(m)|f(0)$. $f(0-m)|f(0)-f(m)\implies f(-m)|f(m)\implies f(m)=f(-m)$. $f(m+n)=f(m-(n))|f(m)-f(-n)=f(m)-f(n)$. Since $f(n)\ge f(m),f(n)>f(n)-f(m)\ge f(m+n)$. Again, $f(n)=f(m+n-m)|f(m+n)-f(m)$. If $f(m)>f(m+n),f(m)>f(m)-f(m+n)\ge f(n)\ge f(m)$, contradiction. If $f(m)<f(m+n),f(m+n)>f(m+n)-f(m)\ge f(n)>f(m+n)$, contradiction! Thus, only $f(m+n)=f(m)$ is possible. Then $f(m)=f(m+n-n)|f(m+n)-f(n)=f(m)-f(n)\implies f(m)|f(n)$.

Let $P(m,n)$ denote the assertion. $P(2k,k)$ gives $f(k)\mid f(2k)$. Then $P((i+1)k,ik)$ gives $f((i+1)k)\equiv f(ik)\pmod f(k)$ so inductively $f(k)\mid f(nk)$ for all integers $n$. This means that $f(k)\mid f(-k) \mid f(k)$ so $f(k)=f(-k)$. $~$ Note that $f(m)|f(0)$ for all $m$ so $f$ is bounded. Suppose $f(m)>f(n)$ then $P(m-n,-n)$ gives $f(m)=f(-m)\mid f(m-n)-f(n)$ so either $f(m-n)>f(n)$ in which case \[f(n)<f(m)\le f(m-n)-f(n)\implies 2f(n)<f(m-n)\]or $f(m-n)<f(n)$, in which case $f(m)\le f(n)-f(m-n)<f(n)$, absurd. Thus, $2f(n)<f(m-n)$. $~$ Let $P(x)$ denote the statement that $xf(n)<f(m-n)$. Then \begin{align*} P(x) &\implies xf(n) < f(m-n) \\ &\implies xf(n) < f(m-n) < f(m)-f(n) \\ &\implies (x+1)f(n) < f(m) \le f(m-n) - f(n) \\ &\implies (x+2)f(n) < f(m-n) \\ &\implies P(x+2) \end{align*}which is surely impossible. Thus, $f(m-n)=f(n)$. Thus, $f(n)\mid f(m)-f(n)$ and we're done.

We denote the given condition with $P(m,n)$.Let $a$ and $b$ be integers such that $f(a)\leq f(b)$. We'll prove either $f(a)=f(b)$, or $f(a-b)=f(a)$. Assume that they are both false. We know from $P(a,b)$ that $f(a-b)|f(a)-f(b)$, so $f(a-b)\leq|f(a)-f(b)|=f(b)-f(a) \iff f(b)\geq f(a-b) + f(a)$. From $P(a, a-b)$ we get that $f(b)|f(a)-f(a-b)$, so $f(b)\leq|f(a)-f(a-b)|$. Therefore, $f(a-b)+f(a)\leq f(b) \leq |f(a)-f(a-b)|$, which is clearly impossible, because $f$ attains only positive values, so the assumption is wrong. Now the conclusion obviously follows from $P(a,b)$.

Pretty good problem but misplaced for p5 definitely.

Nice problem We get $f(m) | f(0)$, $f(-n) | f(0) +f(n) \implies f(-n) | f(n)$ and $f(0 -(-n)) | f(0) - f(n) \implies f(-n) = f(n)$. Next we have $f(m-n) | f(m) - f(n)$, so if $f(m) \neq f(n)$, we have $f(m-n) \leq |f(m) - f(n)|$. FTSOC Consider $f(n) < f(m)$ such that $f(n) \nmid f(m)$ We have $f(m-n) \leq |f(m) - f(n)|$ and $f((m - n) - (-n) | f(m-n) - f(n)$. If $f(m-n) \neq f(n)$, we have $f(m) \leq |f(m-n) - f(n)|$. If $f(m-n) > f(n)$, we have $f(m) + f(n) \leq f(n)$, impossible. We have a similar case for $f(m-n) < f(n)$, forcing $f(m-n) = f(n)$. But then $f(m - (m-n)) | f(m) - f(n) \implies f(n) | f(m)$, a contradiction Thus $f(m) \leq f(n) \implies f(m) | f(n), \blacksquare$

Claim: The image of $f$ is the divisors of $f(0)$. Proof. Since $f(m) \mid f(m) - f(0)$ for all $m$ it follows that $f(m) \mid f(0)$. $\blacksquare$ As such, let the divisors of $c$ be $1 = d_1 < d_2 < \dots < d_{k-1} < d_k = f(0)$. Claim: For all $n$, $d_i \mid d_n$ for $i \le n$. Proof. The base case of $k$ follows immediately. We now induct downwards. Suppose that it holds for $i + 1, \dots, k$. Take minimal $t \ne 0$ such that $f(n) = d_i$. Suppose that $f(m) < f(n)$. Then $d_i \mid f(m + n) - f(n)$. $f(m + n)$ is also a divisor of $f(0)$. Since $f(m) \not\equiv 0 \pmod{d_i}$, it follows that $f(m + n)$ can't be a divisor larger than $d_n$, which implies $f(m + n) = f(m)$. As such, it follows that $f(m) \mid f(m + n) - f(n)$, so $f(m) \mid f(n)$ follows. $\blacksquare$

Let $P(m, n)$ be given assertion. $P(n, 0)$ gives $f(n) \mid f(0)$. Then $P(0, n)$ gives $f(-n) \mid f(0) - f(n)$, so $f(-n) \mid f(n)$. Similarly we have $f(n) \mid f(-n)$. Therefore $f(n) = f(-n)$. Now let $m ,n$ be positive integers such that $f(m) \le f(n)$. Let $a = f(m),b = f(n), c= f(n - m)$. Then we have $a \mid b - c$ and $c \mid b - a$. Since $f(n) = f(-n) \mid f(m - n) - f(m)$, so $b \mid c - a$. If $b = \max(a, b ,c)$, then since $b \mid c - a$, so $c = a$. Thus $a \mid b$. So $c = \max(a, b, c)$. Therefore $a = b$. So in all case $f(m) \mid f(n)$. Thus we're done. $\blacksquare$

Plugging in $0$ for $m, n$ respectively yields \begin{align*} f(m) &\mid f(m) - f(0) \\ f(-n) &\mid f(0) - f(n). \end{align*}In particular, $f(m) \mid f(0)$ for every $m$, which implies $f(-n) \mid f(n)$. Symmetrically, this forces $f(n) = f(-n)$. Now assume $f(m) \geq f(n)$ and consider the relations \begin{align*} f(m) &\mid f(m+n) - f(n) \\ f(n) &\mid f(m+n) - f(m) \\ f(m+n) &\mid f(m) - f(n). \end{align*}Let $d = \gcd(f(m), f(n))$, and let $f(m) = dx$, $f(n) = dy$, $f(m+n) = dz$, where $\gcd(x, y) = 1$. Then we have the divisibility relations $x \mid z-y$, $y \mid z-x$, and $z \mid x-y$. In particular, $xy \mid z-y-x$ as $\gcd(x, y) = 1$. On the other hand, as $z \mid x-y$, it follows that $|z-y-x| < x+y$ and thus $xy < x+y$. As $(x, y) = (2, 2)$ is invalid, this forces $x=1$ or $y = 1$, which is clearly enough.

Let $P(m,n)$ be the assertion of $f(m-n)\mid f(m)-f(n)$. First, $P(a,0)$ asserts that $f(a)\mid f(a)-f(0) \implies f(a)\mid f(0)$ for all $a \in \mathbb{Z}$. Lemma 1. We have $f(a)=f(-a)$ for all $a\in \mathbb{Z}$. Note that for every $n\in \mathbb{Z}$, $P(n+ik+k,n+ik)$ asserts that $f(n+ik+k)\equiv f(n+ik) \pmod {f(k)}$. This means, $\cdots\equiv f(n-2k)\equiv f(n-k) \equiv f(n)\equiv f(n+k)\equiv f(n+2k)\equiv \cdots \pmod{f(k)}$. In other word, if there is positive integers $k,a$, and $b$ with $k\mid a-b$, then we have $f(k)\mid f(a)-f(b)$. Now, if there is an integer $t$ such that $a=kt$, taking $b=k$, we have $k\mid a-b$. This implies $f(k)\mid f(kt)-f(k) \implies f(k)\mid f(kt)$. Hence, $f(k)\mid f(kt)$ for all integers $t$. Hence, we now have $f(k)\mid f(-k)$ and $f(-k)\mid f(k)$. This forces $f(k)=f(-k)$ for all integer $k$. $\blacksquare$ For the sake of contradiction, assume that there is $a,b\in \mathbb{Z}$ such that $f(b)>f(a)$, but $f(a)\nmid f(b)$. Now, $P(a,a-b)$ and $P(b,b-a)$ assert that $f(b)\mid f(a)-f(a-b)\implies f(b)\mid f(a)+f(b)-f(a-b)$ and $f(a)\mid f(a)+f(b)-f(a-b)$, consecutively. This implies $lcm(f(a),f(b))\mid f(a)+f(b)-f(a-b)$. However, by $P(a,b)$, we have $f(a-b)\mid f(b)-f(a) \implies f(a-b)\le f(b)-f(a) \implies f(a)+f(b)-f(a-b)\ge 2f(a)>0$. From $f(a)\nmid f(b)$, we have $lcm(f(a),f(b))\ge 2f(b)$. This implies $2f(b)\le lcm(f(a),f(b)) \le f(a)+f(b)-f(a-b) \implies f(b)\le f(a)-f(a-b) <f(a) \implies f(b)<f(a)$ which is a contradiction. Hence, it is true that for all $m,n \in \mathbb{Z}$ with $f(m)\le f(n)$, we must have $f(m)\mid f(n)$.

Substituting $(x,0)$, $(-x,0)$, $(0,x)$, $(0,-x)$ tells us $f(x)=f(-x)$, so we only need to focus on $x \in \mathbb{Z}_{\ge 0}$. Suppose we have $a+b+c=0$ for integers $a$, $b$, $c$. Then we have \[f(a) \mid f(-b) - f(-c) = f(b)-f(c)\] and cyclic permutations, from which it follows that the minimum two of $f(a),f(b),f(c)$ must be equal. Substituting back into our system, they must both divide the maximum, as desired. $\blacksquare$

Taking $P(m,0)$ gives us $f(m) \mid f(m) - f(0) \implies f(m) \mid f(0)$ for all $m$. Taking $P(0,m)$ and $P(0,-m)$ implies that $f$ is even. Claim: We have $f(a) \mid f(b)$ whenever $a \mid b$. Proof: Since $f$ is even, assume WLOG that $a$ and $b$ are positive. Then, we have $f(a) \mid f(ak) - f(a(k-1))$, so adding these divisibilities together over $k=1, 2, \dots, \tfrac{b}{a}$ gives us the desired claim. So, $f(1) \mid f(m)$ for all $m$. From hereon, assume WLOG that $f(1) = f(-1) = 1$. Claim: If $f(a) \neq 1$ and $f(b) \neq 1$, then $f(a-b) \neq 1$. Proof: Assume FTSOC that $f(a-b) = f(b-a) = 1$. From $P(b,b-a)$ and $P(a, a-b)$ we have \[\begin{cases} f(a) & \mid f(b) - 1 \\ f(b) & \mid f(a) - 1. \end{cases}\]Since both $f(a)$ and $f(b)$ are greater than $1$, this implies that $f(a) > f(b)$ and $f(b) > f(a)$, giving us a contradiction. So, by the Euclidean algorithm, the values of $f$ which map to something not equal to $1$ are all multiples of $k$ for some integer $k \geq 2$ (here we use our first claim as well). We now finish the problem with induction on $f(0)$: The base case of $f(0) = 1$ is trivial. We will show that the problem statement is true for $f(0) = C$ if it is true for $f(0)$ equal to every proper divisor of $C$. If $f(x) = 0$ for all positive integers $x$, we are automatically done. Otherwise, from our previous claim, there exists an integer $k \geq 2$ for which $f(x) \neq 1$ if and only if $k \mid x$. From our first claim, for any multiple $x$ of $k$, $f(k) \mid f(x)$. Therefore, the function \[g(x) = \frac{f(kx)}{f(k)}\]satisfies the same conditions set on $f$ by the problem statement. By assumption, $f(k) > 1$, so $g(0)$ is a proper divisor of $f(0)$, thus the inductive assumption holds on $g$.

Is this correct? Let $P(m,n)$ be the assertion $f(m-n)\mid f(m)-f(n)$ $P(n,0)\implies f(n)\mid f(n)-f(0)\implies f(n)\mid f(0)$ $P(0,n)\implies f(-n)\mid f(0)-f(n)\implies f(-n)\mid f(n)$ $P(0,-n)\implies f(n)\mid f(0)+f(-n)\implies f(n)=f(-n)\implies f$ is even $P(m,-n)\implies f(m+n)\mid f(m)-f(n)$ $P(m,m+n)\implies f(n)\mid f(m)-f(m+n)$ $P(n,m+n)\implies f(m)\mid f(n)-f(m+n)$ Let $f(m)=a, f(n)=b$ and $f(m+n)=c$, then the last 3 conditions are equivalent to $$a\mid b-c,\quad b\mid a-c,\quad c\mid a-b$$WLOG we take $a$ to be maximal so $a>\lvert b-c\rvert$ but we also have $a\mid b-c\implies a\leq b-c$. This gives us two of $f(m), f(n), f(m+n)$ equal and dividing the third. If $f(m+n)=f(m)$ we get $f(m)\mid f(n)$ and if $f(m+n)=f(n)$ we get $f(n)\mid f(m)$, giving us our desired result.

Does this work? Let $P(m,n)$ denote the given assertion. Claim: $f(n)=f(-n) \forall n \in \mathbb{Z} \iff f-\text{even}$ Proof: $P(n,0) \implies f(n) \mid f(n)-f(0) \implies f(n) \mid -f(0) \implies f(n) \mid f(0), n \rightarrow -n \implies \boxed{ f(-n) \mid f(0)}$ $...(*)$ $P(0,n) \implies f(n) \mid f(0)-f(n)$ , combining with $(*)$ we get: $f(n) \mid f(-n) , n \rightarrow -n \implies f(-n) \mid f(n) \therefore f(-n) \mid f(n) \mid f(-n) \implies f(n)=f(-n) \iff f-\text{even}$ $\square$ Now by taking $P(n,-1) , P(n+1,n) \text{and} P(n+1,1)$ and combining with $f-\text{even}$ we get: $\boxed{f(n+1) \mid f(n)-f(1)}$ $...(1) $ $\boxed{f(1) \mid f(n+1)-f(n)}$ $...(2)$ $\boxed{f(n)\mid f(n+1)-f(1)}$ $...(3)$ Now we seperate the problem into $2$ cases. Case 1. $f(n+1) \geq f(n)$. From $(1) \implies f(n+1) \mid f(n)-f(1) \text{but since} LHS > RHS \implies RHS=0 \implies f(n)-f(1)=0 \implies f(n)=f(1) $ $(LHS>RHS \because f(n+1) \geq f(n) )$ By $(3) \implies f(n)\mid f(n+1)-f(1) \implies f(n) \mid f(n+1)-f(n) \implies \boxed{f(n) \mid f(n+1)}$ So $\forall m$ s.t $f(n) \leq f(m) \implies f(n) \mid f(m)$ $\square$ Case 2. $f(n) \geq f(n+1)$ We begin by making the following claim Claim: $f(1) \leq f(n) $ Proof: From $(2) \implies f(1) \mid f(n+1)-f(n)$ but since $f(n) \geq f(n+1)$ we get: $f(1) \mid f(n)-f(n+1) \implies f(1) \leq f(n)-f(n+1) so f(1) \leq f(1)+f(n+1) \leq f(1)+f(n) \leq f(n) \implies \boxed{f(1) \leq f(n)}$ $\square$ Subcase 2.1 $f(n+1) \geq f(1)$ From $(3) \implies f(n) \mid f(n+1)-f(1)$ but $LHS>RHS \implies RHS=0 \implies f(n+1)=f(1)$ From $(1) \implies f(n+1) \mid f(n)-f(1) \implies f(n+1) \mid f(n)-f(n+1) \implies \boxed{f(n+1) \mid f(n)}$ So $\forall m$ s.t $f(m) \leq f(n) \implies f(m) \mid f(n)$ $\square$ Subcase 2.2 $f(n+1) <f(1)$ From $(3) \implies f(n) \mid f(n+1)-f(1)$ but $LHS >RHS \implies RHS=0 \implies f(n+1)-f(1)=0 \implies f(n+1)=1 \rightarrow \leftarrow$ $\blacksquare$ 100th post

Plugging in $n=0$ gives $f(m)|f(0)$. Plugging in $m=0$ gives $f(-n)|f(n)$, so $f(n)=f(-n)$. FTSOC, assume that $f(m)\le f(n)$ and $f(m)\not |f(n)$. We see that $f(m+n) | (f(m)-f(n))$. If $f(m+n)\ge f(n)$, then $f(m)=f(n)$. Therefore, we assume $f(m+n)<f(n)$. Then, $f(n)|(f(m+n)-f(m))$, so $f(m+n)=f(m)$. Then, since $f(m)=f(m+n) | (f(m)-f(n))$, $f(m)|f(n)$, as desired.

Let $P(m,n)$ denote the assertion $f(m-n)|f(m)-f(n)$ Take some $f(x)>f(y)$ $P(y, y-x)$ gives us $f(x)|f(y)-f(y-x)$, but we have that $f(y)-f(y-x) < f(y) < f(x)$, which gives us that $f(y)=f(y-x)$ $P(y,x)$ gives that $f(y-x)=f(y)|f(y)-f(x)$, which means that $f(y)|f(x)$, giving what is desired.

Note that $f(m) \mid f(m)-f(0)$ implies $f(m) \mid f(0)$. Now $f(n) \mid f(0)-f(-n)$ implies $f(n) \mid f(-n)$. Taking $-n$ instead, we get $f(-n)\mid f(n)$ and thus $f(n)=f(-n)$ (since $f>0$). Take $a, b$ such that $f(a)<f(b)$. Taking $m=b, n=-a$ we get $f(a+b) \mid f(b)-f(a)$ which implies $f(a+b)<f(b)$. Now taking $m=a+b, n=a$ we get $f(b) \mid f(a+b)-f(a)$ which implies $f(a+b)=f(a)$ since $|f(a+b)-f(a)|<f(b)$. Finally, $$f(a) \mid f(a+b)-f(b)=f(a)-f(b) \implies f(a)\mid f(b),$$as desired.