Alternative solution:

I think this can be done in a simpler way Let $p_1,p_2, \cdots, p_r$ be the only primes dividing $f(c)$, for all natural number $c$. Let $f(S) = \prod_{i=1}^{r}(p_i)^{\alpha_i}$, where $\alpha_i \ge 0$ Now we will show that the sequence of natural numbers $\left \langle T_n \right \rangle _{n \ge 1 } $ such that $f(T_n + S) = f(S)$ is unbounded from above. Indeed if for all $n \ge 1$ we let $T_n = \prod_{i=1}^{r}(p_i)^{\beta_i}$, such that $\beta_i > \alpha_i$ we have $ (T_n + S) - S \mid f(T_n + S) - f(S)$ and its easy to see that $f(T_n + S) $ and $f(S)$ has same canonical factorization., hence are equal. and take any natural number $N$, so we have $T_n + S - N \mid f(S) - f(N)$, if $T_n + S - N \le f(S) - f(N)$ or we have $T_n \le f(S) - f(N) + N - S$ for all $ n \ge 1$, which is a contradiction, since the sequence is unbounded from above. So we must have $f(S) = f(N)$, so the function is constant, which proves the statement of the problem.

Much easier way: Let f(0)=c We have p|f(p)-c for all p, so f(p)=kp+c for some integer k. Now assume by contradiction that there are finitely many primes that divide f(a) for some positive integer a. Let their product be P. We have $f(P^ac)=kP^ac+c=c(kP^a+1) \forall a \in \mathbb{N}$. Obviously, $kP^a+1$ is not divisible by any of the primes in our product, so k must be 0. Thus, $f(P^ac)=f(0) \forall a \in \mathbb{N}$. This means that $P^ac-x|f(0)-f(x) \forall a,x \in \mathbb{N}$. However, if we fix x, we can choose an a such that $P^ac-x>|f(0)-f(x)|$, so $f(x)=f(0) \forall x \in \mathbb{N}$. However, this means that f is constant, a contradiction. Thus, the problem statement is true.

Quote: so k must be 0. Why?

Zhero wrote: Quote: so k must be 0. Why? Because if k is not 0, $f(P^ac)=c(kP^a+1)$ is divisible by some prime not included in our product P. Unless c=0, but this is ok because if c=0 then p|f(p) for all p. EDIT: Ultimately, this solution is identical to that of the same condition with polynomials, except we need to do a little extra for this problem, because the polynomial version would be finished after the step where $f(P^ac)=c(kP^a+1)$.

Okay, I see. Thank you for clearing that up. However, the problem specifies that the domain and range of $f$ is the set of positive integers, so you're not actually allowed to plug in 0.

Zhero wrote: Okay, I see. Thank you for clearing that up. However, the problem specifies that the domain and range of $f$ is the set of positive integers, so you're not actually allowed to plug in 0. Oops sorry. But i think we can make a similar argument using f(1)=c and $f(P^ac+1)=c(kP^a+1)$.

Alternative solution: So assume the only primes dividing some $f(n)$ are $p_1, p_2, \cdots, p_k$, and let $\gcd(f(1), f(2), f(3), \cdots) = p_1^{e_1}p_2^{e_2}p_3^{e_3}\cdots p_k^{e_k}$. (It is possible for some $e_i = 0$). Note that $p_i^{e_i+1} \vert f(x+p_i^{e_i+1}) - f(x)$. Because the $\gcd = p_1^{e_1}p_2^{e_2}p_3^{e_3}\cdots p_k^{e_k}$, we can find a $x_i$ so that $p_1^{e_i+1} \not\vert f(x_i)$ for any $1 \le i \le k$. Now consider the set of congruences $x \equiv x_i \pmod{p_i^{e_i+1}}$ for $1 \le i \le k$. By the Chinese Remainder Theorem, this gives an arithmetic progression of solutions for $x$. Call the set of elements in this progression $S$. Then for all $x \in S$ we can write $f(x) = p_1^{e_1}p_2^{e_2}p_3^{e_3}\cdots p_k^{e_k} \cdot q$. By our construction of $x$, $p_i \not\vert q$ for all $i$. But the since the only primes dividing some $f(n)$ are the $p_i$, we get that $q = 1$. So all elements $x \in S$, $f(x) = p_1^{e_1}p_2^{e_2}p_3^{e_3}\cdots p_k^{e_k}$. Since $f$ is nonconstant, we can take a $y$ so that $f(y) \not= p_1^{e_1}p_2^{e_2}p_3^{e_3}\cdots p_k^{e_k}$, and then $x-y \vert f(x) - f(y) \implies x-y \vert p_1^{e_1}p_2^{e_2}p_3^{e_3}\cdots p_k^{e_k}-f(y)$ when $x \in S$. But $x-y$ can get arbitrarily large, contradiction.

to filetmiqnon821 the function maps from positive integers to itself............................how can u take 0.

Suppose the problem is false. Then there exists a $k \in \mathbb{N}$ such that there exists distinct primes $p_1$, ..., $p_k$ such that there exists functions $g_i : \mathbb{N} \mapsto \mathbb{N}_0$ such that $f(a)=\prod_{i=1}^k p_i^{g_i(a)}$, $\forall a \in \mathbb{N}$. Now, take $b = M \left( \prod_{i=1}^k p_i^{g_i(a)+1} \right) + a$, for any $M \in \mathbb{N}$. Then we have $X = \left( \prod_{i=1}^k p_i^{g_i(a)+1} \right)$ $|$ $b-a$ $|$ $f(b)-f(a) = \prod_{i=1}^k p_i^{g_i(b)} - \prod_{i=1}^k p_i^{g_i(a)}$. Notice that $v_{p_i}(X) = g_i(a)+1$ for all $i \le k$, and therefore $v_{p_i}(f(b)-f(a)) \ge g_i(a)+1 = v_{p_i}(f(a))+1$. So we must have $v_{p_i}(f(a))=v_{p_i}(f(b))$ for all $i \le k$. But since no other prime divides $f(x)$ (for all $x$), then we have $f(a)=f(b)$ for any $M$. Notice that we can choose $M$ as big as we want, and they all generate different $b$'s. Therefore, since $f$ is not constant, we can take $c$ such that $f(c) \neq f(a)$. For any $M$ we must have $b-c | f(b)-f(c) = f(a)-f(c)$ and so $|b-c| \le f(a)-f(c)$ and so $b \le f(a)-f(c)+c$. But if we choose a $M$ that is very very gigantic, this will be impossible. So we're done.

Very easy for $N3$,but still very nice.Let f(1)=p1^a1*p2^a2*...pn^an.Now,if we pick a number of form $S*k+1$ where $k$ is an arbirtary positive integer and $S=f(1)*p1*p2...*pn$,then $f(S*k+1)$ is equal to $f(1)$,so we have infinity many integers $c$ such that $f(c)=f(1)$,so we can pick an integer $n$ such that $f(n)$ isn't equal $f(1)$(function isn't constant),so we have $n$-very big $c$ divides $f(n)-f(1)$,but this is a contradiction cause we can pick an arbirtary large $c$.

We adapt the proof of Schur's theorem from Problems from the book. Again, let $g(x) = f(x+1)$, so the domain of $g$ is nonnegative integers. Now if $g(0) = 0$ the question is trivial, because $f(p)$ is a multiple of $p$ for all primes $p$. Now consider the function \[ h(x) = \frac{g(xg(0))}{g(0)} \]. Now this is integer valued, because \[ \frac{g(xg(0)) - g(0)}{xg(0)} \] is an integer, implying this result. Now observe that it is sufficient to show that $h$ has infinitely many prime divisors. But this is easy. Assume that the largest prime divisor dividing any element of $f$ is $P$, and now observe that $h(xP!) - h(0) \equiv 0 \pmod{P!}$, so $h(xP!) \equiv 1 \pmod{P!}$. Hence, $h(xP!) = 1$ for all integers $x$, but then $h(xP!) - h(2) > xP! - 2$, which is false for significantly large $x$, so we are done.

Lemma: For a fixed $c$ we can only have finitely many $x\in \mathbb{N}$ s.t. $f(x)=c$. Proof: Going by contradiction, let $M$ be the (infinite) set of positive integers with $f(x)=c$. As $f$ is nonconstant, there exists a positive integer $n$ s.t. $f(n)\ne c$. For any $x\in M$ we have that $x-n$ divides $c-f(n)$. As $x-n$ is unbounded, we get that $f(n)=c$, contradiction. Lemma: Given a positive integer $n$ such that $f(n)\ne 1$, we can find a positive integer $m$ such that $f(n)|f(m)$ and $f(m)$ has at least one prime divisor which cannot be found in $f(n)$. Proof: Let $f(n)=p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$ and denote $a=p_1^{\alpha_1+1}p_2^{\alpha_2+1}...p_k^{\alpha_k+1}$. Then $a|f(n+a)-f(n)$ and $a|f(n+ta)-f(n+(t-1)a)$, hence (by summing) $a|f(n+ta)-f(n)$ for all positive integers $n$. By the previous lemma, $f(n+ta)$ cannot be equal to $f(n)$ for all values of $t$. Let $k$ be a positive integer such that $f(n+ka)\ne f(a)$. As $a|f(n+ka)-f(n)$, we infer that $f(n)|a|f(n+ka)-f(n)$, hence $f(n+ka)=f(n)r$ with $r\ne 1$. We get $p_1...p_k|r-1$, so $r$ has a prime divisor which is not among $p_1,...,p_k$, i.e. $f(n+ka)$ has all the prime divisors of $f(n)$ and at least one more. As $f$ is nonconstant, we have a positive integer $n$ such that $f(n)\ne 1$. Starting with $n$ and applying the second lemma repeatedly, we find positive integers $m$ such that the number of prime divisors of $f(m)$ is as big as we want, whence the conclusion.

Suppose that, for finite $n,$ $ p_1,...,p_n$ are prime divisors of $f.$ Let $f(1)=\prod_{i=1}^np_i^{\alpha_i}$ and let $u$ be such that $v_{p_i}(u)>v_{p_i} (f(1)).$ For any $v \in \mathbb N$ we have $uv|f(uv+1)-f(1).$ Consider any $p_i.$ Since $v_{p_i}(u)>v_{p_i}(f(1))$ and $v_{p_i}(u)<v_{p_i}(f(uv+1)-f(1))$ we clearly must have $$v_{p_i}(f(uv+1))=v_{p_i}(f(1))$$Now since, by our assumption, $p_i$'s are only possible prime divisors of $f(uv+1)$ we must have $f(uv+1)=f(1), \forall v \in \mathbb N.$ Now take any $m \in \mathbb N.$ We must have $m-(uv+1)|f(m)-f(uv+1)=f(m)-f(1).$ So now as $v$ runs to infinity we get $|LHS|>|RHS|$ hence our assumption is wrong, so we are done.

For the sake of contradiction suppose that only the primes in the set $S = \{p_1, p_2, \cdots, p_k\}$ divide $f(n)$ for any positive integer $n$. Define \[G \overset{\text{def}}{=} \gcd(f(1), f(2), f(3), \cdots) = p_1^{e_1}p_2^{e_2}p_3^{e_3}\cdots p_k^{e_k}\]Define the new function $g: \mathbb{N} \rightarrow \mathbb{N}$ such that \[g(n) \overset{\text{def}}{=} \frac{f(n)}{G}\]Lemma: If $p \in S$ and $v_p(G)=e$, then $g(n+p^{e+1}) \equiv g(n) \pmod{p}$ Proof. The problem condition gives $p^{e+1} \mid f(n + p^{e+1}) - f(n)$. Hence \[g(n+p^{e+1}) - g(n) = \frac{f(n+p^{e+1}) - f(n)}{G} \equiv 0 \pmod{p}\]So the period of $f(n) \pmod{p}$ is $p^{e+1}$. $\blacksquare$ The definition of $G$ gives that for each $p_i \in S$, there exists $n_i$ such that $\gcd(f(n_i), p_i) = 1$. By the Chinese Remainder Theorem, there exist infinitely many $N$ such that \begin{align*} N &\equiv n_1 \pmod{p_1^{e_1+1}}\\ N &\equiv n_2 \pmod{p_2^{e_2+1}}\\ &\cdots \\ N &\equiv n_k \pmod{p_k^{e_k+1}} \end{align*}This definition of $N$ means $\gcd(g(n), p) = 1$ for every $p \in S$. But the only prime factors of $g$ are in $S$. So $g(N)=1$ and $f(N) = G$. Fix an arbitrary integer $a$ and let $N$ be arbitrarily large. The problem condition gives \[\lim_{N \rightarrow \infty} \frac{f(N) - f(a)}{N - a} = \lim_{N \rightarrow \infty}\frac{G-f(a)}{N-a} \in \mathbb{Z} \implies f(a)=G\]which means $f(n)$ is constant, a contradiction.

Please check my proof

Absolutely Kayak wrote: Please check my proof

Perfect

Assume that the only primes which divide an element of $f({\mathbb N})$ are the primes $p_1, p_2, \dots, p_n$; we will prove $f$ is constant. Let \[ f(1) = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k} \](allowing $e_i = 0$) and define \[ N = p_1^{e_1+1} p_2^{e_2+1} \dots p_k^{e_k+1}. \] Claim: The numbers $f(N+1)$, $f(2N+1)$, $f(3N+1)$, \dots all divide $f(1)$. Proof. We note $f(N+1) \equiv f(1) \pmod N$, so the exponent of each prime in $f(N+1)$ is at most $e_i$. Since $f(N+1)$ are only allowed to have $p_i$ as prime factors, that implies $f(N+1)$ divides $f(1)$. The proof for $f(2N+1)$, $f(3N+1)$, and so on are identical. $\blacksquare$ But there are finitely many divisors of $f(1)$, so some divisor appears infinitely often in the sequence $f(N+1)$, $f(2N+1)$, ...(pigeonhole principle). Call that divisor $d$. Now consider any fixed $n$. We have $m-n \mid d-f(n)$ for any $m$ with $f(m) = d$. Since $m$ can be arbitrarily large, that means $f(n) = d$. Remark: Philosophy remark: the condition can be interpreted as saying that if $a \equiv b \pmod n$ then $f(a) \equiv f(b) \pmod n$. In that sense this number theory problem uses neither the addition nor multiplication operation. Consequently, for any given prime $p$ it seems impossible to show that $p \mid f(c)$ for some $c$. So the use of contradiction here is substantial.

First of all, note that $k\mid f(a+kt)-f(a)$ for all positive integers $a,k,t$. We also must have a constant $x$ for which $f(x)\ge 2$. This means that there must be a prime $p$ that divides $f(c)$ for a positive integer $c$. For the sake of contradiction, let $p_1,p_2,...,p_k$ be the complete list of primes that divides $f$. Let $p_i\mid f(a_i)$ for all $i=1,2,...,k$. Now, choose positive integer $m\equiv a_i \pmod{p_i}$ for all $i=1,2,...,k$. This number satisfies $p_1p_2...p_k\mid f(m)$. Let $a_i=v_{p_i}(f(m))$ for all $i=1,2,...,k$. Consider the number $L=sp_1^{a_1+1}p_2^{a_2+1}...p_k^{a_k+1}+m$ where $s$ can be any positive integer. Notice that $p_i^{a_i+1}\mid f(L)-f(m)$ for all $i=1,2,...,k$. This implies $v_{p_i}(f(L))=a_i$ for all $i=1,2,...,k$. Because the primes dividing $f$ can only be $p_1,p_2,...,p_k$, this forces $f(L)=f(m)$. Now, note that $(sp_1^{a_1+1}p_2^{a_2+1}...p_k^{a_k+1}+m)-b\mid f(m)-f(b)$ for every positive integer $s$ and $b$. By taking $s$ large enough, we conclude $f(b)=f(m)$ for all positive integer $b$. This leads us to contradiction as $f$ is non-constant. We're done.

Assume the contrary, let $p_1, p_2, \dots, p_k$ be all primes dividing the sequence $f(1), f(2), f(3), \dots$. Let $f(1) = p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ and let $T = p_1^{a_1+1}p_2^{a_2+1}\cdots p_k^{a_k+1}$. Then consider the following claim: Claim: $f(aT + 1) = f(1)$ for all $a \ge 1$. Proof. Note that $T \mid aT + 1 - 1 \mid f(aT + 1) - f(1)$, hence $\nu_{p_i}(T) = a_i+1 \le \nu_{p_i}(f(aT + 1) - f(1))$. If $\nu_{p_i}(f(1)) \neq \nu_{p_i}(f(aT+1))$, then $a_i+1 \le \nu_{p_i}(f(aT+1) - f(1)) = \min(\nu_{p_i}(f(1)), \nu_{p_i}(f(aT+1))) \le \nu_{p_i}(f(1)) = a_i$, a contradiction. Therefore $\nu_{p_i}(f(1)) = \nu_{p_i}(f(aT+1))$ for all $i$, which means $f(1) = f(aT+1)$ for all $a \in \mathbb{N}$. Now take $b$ such that $f(b) \neq f(1)$. Then for large $n$, we have $nT + 1 - b \mid f(nT + 1) - f(b) = f(1) - f(b)$, thus $nT + 1 - b \le |f(1) - f(b)|$, contradicting the fact that $n$ is sufficiently large. Hence $f(b) = f(1)$ for all $b$, contradicting $f$ is nonconstant. So we're done. $\blacksquare$

Otherwise suppose $f$ is $p$-smooth. For primes $2\le q\le p$ take $n_q$ to be the smallest positive integer such that $\nu_q(f(n_q))$ is minimized and call this minimum $m_q.$ Then $f(n_q+k\cdot q^{m_q+1})$ also shares this minimum $\nu_q$ for any positive integer $k.$ Now by chinese remainder theorem there is a solution to the congruence system $z \equiv n_q\pmod{q^{m_q+1}}$ for all $q.$ We then have $\nu_q(f(z))=m_q$ for all $q.$ Now let $\prod_{q\in\mathbb P,q\le p}q^{m_q}=\mathcal L.$ Consider $\frac{f(z)}{\mathcal L}.$ By $\nu_q$ it cannot be divisible by any prime $2\le q\le p.$ By assumption it cannot be divisible by any prime $>p,$ so it must be $1$ so $f(z)=\mathcal L.$ Similarly, we have $f(z+k\mathcal L\cdot p\#)=\mathcal L$ for any positive integer $k,$ where $p\#$ is the primorial of $p.$ Thus $f(x)=\mathcal L$ for infinitely many $x.$ There must exist some $y$ with $f(y)\ne\mathcal L.$ However we must have $x-y\mid \mathcal L-f(y)$ for arbitrarily large $x,$ contradiction. Thus our original assumption is false.

Suppose for the sake of contradiction that the only primes dividing any value of $f$ are $p_1$ through $p_n$. Let $P=f(1)\times\prod p_i$, so that $\nu_{p_i}(P)>\nu_{p_i}(f(1))$ for each $p_i$. $P\mid f(1+P)-f(1)$, so $\nu_{p_i}(f(1))=\nu_{p_i}(f(1+P))$ for each $p_i$. This means that $f(1+P)=f(1)$. Similarly we can prove by induction that $f(1+kP)=f(1)$ for all positive integers $k$. Then for any integer $m$, choose $k$ large such that $|kP+1-m| > |f(1)-f(m)|$. Then $kP+1-m\mid f(kP+1)-f(m)=f(1)-f(m)$, so $f(1)=f(m)$ for all $m$. But $f$ is non-constant, contradiction.

Suppose for contradiction there are only finitely many such primes, say from the set $\{p_1,p_2,\ldots,p_k\}$. We claim that the function has to be constant. Define $P=f(1)\cdot p_1p_2\cdots p_k$. Claim: $f(Pn+1)=f(1)$ for each $n$. Proof. From original condition, $P\mid f(Pn+1)-f(1)$, and since $f(1)\mid P$ we have that $f(1)\mid f(Pn+1)$. Let $f(Pn+1)=xf(1)$. Then, $p_i\mid x-1$ for each $1\leq i\leq k$. Suppose $x\ne 1$. Since $x$ is not divisible by any prime from $\{p_1,p_2,\ldots,p_k\}$, there is some new prime that divides $x$. So there's a new prime which divides $f(Pn+1)$, which is impossible. Hence, $x=1$ is forced.So, $f(Pn+1)=f(1)$, as claimed. We now show that $f(n)=f(1)$ for each $n$. Suppose $f(n)\ne f(1)$. Then, for each positive integer $m$, we have (from our claim) \[Pm+1-n\mid f(Pm+1)-f(n)=f(1)-f(n)\]Since $m$ can be arbitrarily large, it follows that $f(1)-f(n)$ has infinitely many divisors, which is impossible. So, $f(1)=f(n)$ is forced for each $n$. So, the function is constant.

FTSOC that there are finitely many $p$ that divide $f$. Then let $f(1) = k = \prod_i p_i^{e_i}$. Then let $\mathcal{P} = \prod_i p_i^{e_i + 1}$. Then we have(for some arbitrary constant $M$) $M\mathcal P + 1 - 1 \mid f(M\mathcal P + 1) - k$ so it follows that $\nu_{p_i}(f(M\mathcal P + 1)) = \nu_{p_i}(k) \implies f(M\mathcal P + 1) = k$. Then take sufficiently large $M$ and some arbitrary $t$ to get $M\mathcal P + 1 - t \mid k - f(t)$ to get a size contradiction, implying $k = f(t)$ for all $t$. However this is a contradiction as $f$ is nonconstant, done.

Suppose that it is not true. Let, $p_{1},\cdots,p_{n}$ are the prime factors of $f(1),f(2),\cdots$ Then $f(a)=p^{a_{1}}_{1} \cdots p^{a_{k}}_{k}$ Let be some $x_{\mathcal{Q}}=\mathcal{Q}*p^{a_1+1}_1 \cdots p^{a_k+1}_k$ So,that $x_{\mathcal{Q}} \vert (f(a+x)-f(a))$ $\nu_{p_i}{f(a)} < \nu_{p_i}{f(x_{\mathcal{Q}})} $ From that, $\nu_{p_i}{f(a)} = \nu_{p_i}{f(a+x_{\mathcal{Q}})} $ Thus it follows every $\mathcal{Q} \geq 1 $ this identity is true. But, if we plug $P(a+x_{\mathcal{Q}}+1,1)$ Since that $f(a)=f(1)$ Which gives us contradiction.

Solved with surpidism. Assume the contrary. The condition can be translated to $N \mid a - b \implies N \mid f(a) - f(b).$ Take an $N$ such that $\nu_p(N) > \nu_p(f(1))$ for all primes $p \mid f(1).$ Then, \[f(kN+1) \equiv f(1) \pmod N.\] As, $\nu_p(N) \leq \nu_p(f(kN+1) - f(1)),$ forcing $\nu_p(f(kN+1)) = \nu_p(f(1)),$ otherwise \[\nu_p(f(1)) < \nu_p(N) \leq \nu_p(f(kN+1) - f(1) \leq \min(\nu_p(f(kN+1)), \nu_p(f(1)))\] a contradiction. As $f(kN+1)$ and $f(1)$ share the same prime divisors, we get $f(kN+1) = f(1).$ To finish, note that \[kN + 1 - a \mid f(kN+1) - f(a) = f(1) - f(a).\] Taking arbitrarily large $k$ such that $|f(a) - f(1)| < kN+1 - a \leq |f(1) - f(a)|$, a contradiction.

Assume, for the sake of contradiction, that there are only finitely many such primes $p_1, p_2, ... p_n$. Substituting $a = (f(1) p_1 p_2 p_3 \ldots p_n) + 1$ and $b = 1$, we get: $(f(1) p_1 p_2 p_3 \ldots p_n) | f((f(1) p_1 p_2 p_3 \ldots p_n) + 1) - f(1)$ $\implies f((f(1) p_1 p_2 p_3 \ldots p_n) + 1) = k (f(1) p_1 p_2 p_3 \ldots p_n) + f(1) = f(1) (k p_1 p_2 p_3 \ldots p_n + 1)$ which clearly has a prime divisor other than $p_1, p_2, ... p_n$ , a contradiction.

2009 N3 Assume FTSOC otherwise, and let {\(p_1\), \(p_2\), \(\ldots\), \(p_n\)} be the set of primes dividing \(f\). Let \(Q=f(1)\cdot p_1p_2\cdots p_n\). Then we Claim that. \[f(Qn+1)=f(1) .\] Proving that we get that \(m-Qn\mid f(m)-f(1)\) ,we can take \(n\) as it approaches infinity, so \(f(m)=f(1)\) for all \(m\), a contradiction.

Let $P$ be the set of primes dividing $f$. Let $m$ be a value such that $f(m)=\prod_{p_i\in P}p_i^{\alpha_i}$, such that $\alpha_i>0$ for all $i$, now let $N$ be a value such $N=\prod_{p_i\in P}p_i^{\beta_i}$ such that $\beta_i>\alpha_i$ for all $i$. Thus we have that $N\mid f(N+m)-f(m)$ however we can see that $\nu_{p_i}(f(N+m))=\nu_{p_i}(f(m))$, thus we get $f(N+m)=f(m)$, so we get arbitrarily large values that are equal to $f(m)$. Now by taking a sufficently large value $k$ such that $f(k)=f(m)$ we can get for values below a certain constant, are constant and by using increasingly large values of $k$ we get the function must be constant.

Easy FTSOC, assume there are finite number of primes $p_1,p_2,...p_n$ that divide any $f(k)$ Consider $a=kn$ and $b=mn$ and let n be such that $GCD ( n, p_i) = 1$ for all $ i $ SInce $n(k-m) \mid f(nk) - f(n) $ and due to our above condition we will have GCD$ ( f(kn) , f(mn) ) =1 $ But since our number of primes are limited we have a contradiction

Please contact westskigamer@gmail.com if there is an error with my solution for cash bounties by 3/18/2025.

I have the exact same solution as mathocean97

Assume contrary. Let $p_1,\dots,p_k$ be the set of prime divisors of $f$. Let $f(1)=p_1^{x_1}\dots p_k^{x_k}$ for $x_i\geq 0$. For $m_i>x_i$, we have $p_1^{m_1}\dots p_k^{m_k}|f(p_1^{m_1}\dots p_k^{m_k}n+1)-f(1)$ and $p_i^{x_i+1}|f(p_1^{m_1}\dots p_k^{m_k}n+1)-p_1^{x_1}\dots p_k^{x_k}$. Thus, $\nu_{p_i}(f(p_1^{m_1}\dots p_k^{m_k}n+1))=x_i$ which implies $f(p_1^{m_1}\dots p_k^{m_k}n+1)=f(1)$. Pick some $x$ and $n\gg f(x)$ in order to have \[p_1^{m_1}\dots p_k^{m_k}n+1-x|f(p_1^{m_1}\dots p_k^{m_k}n+1)-f(x)=f(1)-f(x)\]$f(1)-f(x)$ is a constant while $p_1^{m_1}\dots p_k^{m_k}n+1-x$ increases hence $f(x)=f(1)$ which gives a contradiction since $f$ is non-constant as desired.$\blacksquare$