Setting $m = f(n)$, we get $f(n)^2 + f(n) \mid f(n)f(f(n)) + n \Rightarrow f(n) \mid n \Rightarrow f(n) \leq n \; \forall \; n \in \mathbb{N}$. This implies $f(1) = 1$. Setting $m=n$, we get $n^2 + f(n) \mid nf(n) + n \Rightarrow n^2+f(n) \leq nf(n) + n \Rightarrow n^2 - n \leq (n-1)f(n)$ This implies $n \leq f(n) \;\forall \; n \geq 2 \Rightarrow f(n) = n \; \forall n \in \mathbb{N}$.

Let $m=n$, then \[m^2+f(m)|m(f(m)+1)\Rightarrow f(m)=\frac{k_m m^2-m}{m-k_m}\] for some $k_m\in\mathbb{Z}^+$, so $1<k_m<m$ and thus $f(m)\leq m^3-m^2-m$, as with $k$ growing the numerator grows and the denominator shrinks, for $m=2$, this gives us $k=1$ and $f(2)=2$. Now use $(2,m)$, then for some $l_m\in\mathbb{Z}^+$ \[4+f(m)|4+m\Rightarrow f(m)=\frac{m+4}{l_m}-4\leq m\] Using $m=1$ and $m=3$ in $4+f(m)|4+m$ we get $f(1)=1$ and $f(3)=3$, so we can assume $m\geq 4$ Now look at the following: \[\frac{m(mk_m-1)}{m-k_m}=f(m)\leq m\] Use $k_m=2$, we get: \[\frac{m(2m-1)}{m-2}-m=\frac{m(m+1)}{2}>0\] This means that only $k_m=1$ can hold, meaning $f(m)=m$.

lyukhson wrote: Let $\mathbb{Z} _{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z} _{>0}\rightarrow \mathbb{Z} _{>0}$ such that \[ m^2 + f(n) \mid mf(m) +n \] for all positive integers $m$ and $n$. we know : $f(p)^{2}+f(p)\mid f(p)f(f(p))+p\Rightarrow f(p)=1\: or\: p$ ($p$ is prime number) $i)$ if we have Infinitely many prime numbers such that $f(p)=p$ so we have: $m^{2}+p\mid mf(m)+p\Rightarrow m^{2}+p\mid mf(m)-m^{2}\Rightarrow mf(m)-m^{2}=0\rightarrow f(m)=m$ $ii)$ if we don't have Infinitely many prime numbers such that $f(p)=p$ ! $m=n=p\rightarrow p^{2}+1\mid 2p$ and it is not true for Infinitely many prime numbers . so $f(m)=m$ is unique solution.

lyukhson wrote: Let $\mathbb{Z} _{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z} _{>0}\rightarrow \mathbb{Z} _{>0}$ such that \[ m^2 + f(n) \mid mf(m) +n \] for all positive integers $m$ and $n$. Let $P(m,n)$ be assertion of $ m^2 + f(n) \mid mf(m) +n $ $P(f(n),n)$ implies $f(n)$ divides $n$ so $f(n) \leq n$ .This implies $f(1)=1$. Now suppose for some $n$ we have $f(n)=n$ (this is possible since $f(1)=1$). We will prove that $f(n+1)=n+1$. Consider: $P(n,n+1)$: Then $ n^2 + f(n+1) \mid nf(n) +n+1 $ from which we obtain $f(n+1) \leq n+1$ $P(n+1,n)$: Then $ (n+1)^2 + n \mid (n+1)f(n+1) +n $ from which we obtain $n+1 \leq f(n+1)$.Together this implies $f(n+1) \leq n+1 \leq f(n+1)$ so $f(n+1)=n+1$. Since $f(1)=1$ we see that $f(n)=n$ for every positive integer $n$.

It's inequality problem. We have that $f(n)\ge n$ and there exists $k\in N$ such that $f(k)\le n+f(1)-1<2k$. So $f(k)=k$. So for all $n\in N$ \[ f(n)=n. \]

Solution with thkim1011: Let $m = n$ and we get $m^2 + f(m) | mf(m) + m$ so we have \[m^2 + f(m) \le mf(m) + m\] which rearranges to $f(m) \ge m$ for all $m \ge 2$. Now let $m = f(n)$ and we get \[f(n)^2 + f(n) | f(f(n))\cdot f(n) + n\] so $f(n) |n$ but $f(n) \ge n$, so the only viable possibility is $f(n) = n$ for $n \ge 2$. For $n = 1$, we have $f(1) | 1$, so $f(1) = 1$. It is easy to check that $f(n) = n$ satisfies the condition.

If we put $m=n$ we get $f(m)\geq m$ for all $m \geq 2$ (*). If we put $m=n=2$ we get $4+f(2)|2f(2)+2$ and thus $f(2) =2$. If we put $m=2$ we get $4+f(n)|4+n$ and thus $f(n)\leq n$ for all $n$ (**). If we use (*) and (**) we get $f(n)=n$ for all $n$.

it was really easy!!! this problem was proposed by Mohsen Jamaali

Nice solution everyone. Was quite easy My first impression was it was very similar to IMO 2004 shortlist problem (N3). Indeed, the solution was very similar. First, I noted that $f(n) = n$ was a solution (as almost always). And I noticed that we need to just prove either $f(1) = 1$ or $f(p) = p$ for all $p$ prime. (*) [Taking $m = n = p$ gives us $p^2 + f(p) \mid p(f(p) + 1)$ Since $p^2 + f(p)$ does not divide $f(p) + 1$ (as its bigger) thus, $p \mid p^2 + f(p)$ i.e. $p \mid f(p)$ Therefore $f(p) \geq p$ If $f(1) = 1$ then take $m = 1$, $n = p$ and we get $1 + f(p) \mid p + 1$. Thus $f(p) = p$ for all p prime. (take inequality) And, if we show $f(p) = p$, then taking m = p we get $p^2 + f(n) \mid p^2 + n$ Thus $p^2 + f(n) \mid (n - f(n))$. Choosing $p$ large enough, we get the quotient tends to 0. (as $n - f(n)$ is fixed but $p$ can be as large as you want). Therefore, $f(n) = n$ for all $n$ in naturals] To prove (*) - it didn't hit me to take $m = f(n)$ as nicetry007 did. (You could probably finish the proof like that) Instead, what I did was the following: Let $p$ be a prime and take $n = p$. Let $q$ be a prime dividing $f(p)$, and take $m = q$. We get: $q^2 + f(p) \mid qf(q) + p$ But $q\mid q^2 + f(p)$ ! Thus, we get $q \mid p$, or $p = q$. So, we have $f(p) = p^i$, where $i$ depends on $p$. Now, take $m = n = p$. We get $p^2 + p^i \mid p^{i+1} + p$. (1) Case 1 : $i \geq 3$ Easy contradiction to (1) as $p^2 | LHS$ but not the $RHS$. Case 2: $ i = 2$ Then we get $2p^2 \mid p^3 + p$ or, $2p \mid p^2 + 1$, this means $p\mid1$. Contradiction. Thus, $f(p) = p$ and we're done.

Plugging $m=n,$ \[m^2 + f(m) \leq mf(m) + m \implies m^2 + f(m) \leq mf(m) + m \implies m \leq f(m).\] Plugging $n=f(m),$ \[f(m) \mid f(m)^2 + f(m) \mid mf(f(m)) + f(m) \implies f(m) \mid m \implies f(m) \leq m.\] Hence, $\boxed{f(m)=m \quad \forall \ m \in \mathbb{N}}.$

Let $m=n$ to get $n^2+f(n)\mid nf(n)+n$, so $n^2+f(n)\le nf(n)+n$, or $f(n)\ge n$. In the case of $n=2$, $4+f(2)\mid 2f(2)+2$, and $\frac{2f(2)+2}{f(2)+4}=2-\frac6{f(2)+4}\in\mathbb{Z}$. Therefore, $f(2)+4\in\{-6,-3,-2,-1,1,2,3,6\}$, and the only solution with a positive $f(2)$ is $f(2)=2$. Let $m=2$ to get $2^2+f(n)\mid2f(2)+n$, so $4+f(n)\le2f(2)+n=4+n$, or $f(n)\le n$. Therefore, $f(n)=n$ is the only solution, and we can easily verify that $m^2+n\mid m^2+n$.

Very easy problem. I find 2 solutions (the same as #8 and #11) at once.

$m=2;n=1 \Rightarrow 4+f(1) | 2f(2)+1 $ $m=1;n=2 \Rightarrow 1+f(2)|f(1)+2 \Rightarrow f(1)+1 \geq f(2) \Rightarrow$ $ 4+f(1) \geq \dfrac{2f(2)+1}{2} \Rightarrow 4+f(1)=2f(2)+1$, yielding to $1+f(2) | 2f(2)-1 \Rightarrow f(2)=2;f(1)=1$ $f(1)=1 -> 1+f(n) | n+1 \Rightarrow n \geq f(n); m^2+1 | mf(m)+1 \Rightarrow m \leq f(m)$, and now we have the conclusion: $f(x)=x$

Letting $m=n$, we can conclude that $f(k) \geq k$ for all $k\in \mathbb{N}$. So \[ \frac{mf(m) +n}{m^2 + f(n)} \leq \frac{mf(m)+n}{m^2 + n}\] Note that $\frac{mf(m)+n}{m^2 + n} \rightarrow 1$ as $n$ goes to infinity. But as $ \frac{mf(m) +n}{m^2 + f(n)}$ is always positive integer we have for all large $s$; \[ \frac{mf(m) +s}{m^2 + f(s)} = 1 \implies mf(m) -m^2 = f(s) -s : = c\] From here there is two easy ways to finish the solution The first is to note that $c$ is clearly independent of $m$.That's for all positive integers $l$, $lf(l) - l^2 = c$ , but letting $l > 1$ large enough so that $f(l)-l =c$, we have $l f(l) - l^2 = f(l) -l \implies l c =c \implies c=0$, so we have $mf(m)-m^2 = 0$ for all $m$, whence $f(m) = m$ is the only solution. The second is to use $f(s)=s + c $ for all large $s$, and substitute it with $s > c + 1$, where we get \[ s^2+c+s \mid s^2 + cs + s \implies s^2 +c +s \mid cs-c \implies cs-c =0 \implies c=0 \] The second implication follows from the inequality; $cs-c < (s-1)^2 < s^2 < s^2 +c +s $

Let $P(m,n)$ be the assertion $ m^2+f(n)\mid mf(m)+n $ Let $f(1)=a$ \[P(a,1) \implies a^2+a|af(a)+1\] \[\implies a|1\] \[\implies f(1)=1\] \[P(1,n) \implies 1+f(n)|n+1\] \[\implies f(n) \le n\] \[P(n,n) \implies n^2+f(n)|nf(n)+n\] \[\implies f(n) \ge n\] Thus $\boxed{f(n)=n}$ is the only solution.

I'll post a rather different solution I've found Plugging $m=n$ yields $n^2+f(n) | nf(n)+n$ implying that $f(n) \geq n$ So , we may put $f(n)=n+k$ where $k \geq 0$ is an integer (since the function maps integers to integers) Plugging in $f(n)=n+k$ , the divisibility condition is equivalent to $m^2+n+k|m(m+k)+n ==> m^2+n+k|k(m-1)$ If $k=0$ , then it is equivalent to $m^2+n+k|0$ wich is always true and we get the solution : $f(n)=n$ If $k>0$ , we just chose $n=km+1$ and plug it in $m^2+n+k|k(m-1)$ , in order to get the contradiction .

Plugging in $m=n=2$, we get $4+f(2)\mid 2f(2)+2$. Note $4+f(2)\mid 8+2f(2)$, so it follows $4+f(2)\mid 6$. Thus $f(2)=2$. On the other hand, plugging in $m=2, n=1$, we get $4+f(1)\mid 4+1$, so $f(1)=1$. Finally, plugging in $m=1$, we get $1+f(n)\mid 1+n$, so $f(n)\le n$. On the other hand, plugging $m=n$ gives $m^2+f(m)\le mf(m)+m$, so $f(m)\ge m$. Combining $f(m)\ge m$ and $f(m)\le m$ gives $f(m)=m$ for all $m$, so we are done.

Solved with heheman and blueberryfaygo_55. We claim the answer is $\boxed{f(n)\equiv n}$. It is easy to check that this works. Let $P(m,n)$ be the given assertion. $P(2,2)$ gives $4+f(2)\mid 2f(2)+2$ so $2f(2)+2=a(4+f(2))$ for some $a\in\mathbb{Z}^+$. Since $f(2)$ is positive $a$ must equal $1$ so $f(2)=2$. $P(2,1)$ gives $4+f(1)\mid 5$ so $f(1)=1$. We prove that $f(n)=n$ for all $n\geq 2$ by induction on $n$. Assume the statement for $n=k$. $P(k,k+1)$ gives $k^2+f(k+1)\mid k^2+k+1$ so $k^2+k+1=a\left(k^2+f(k+1)\right)$ for some $a\in\mathbb{Z}^+$. Then $af(k+1)=(1-a)k^2+k+1$, which is positive only when $a=1$. Thus $f(k+1)=k+1$ so we are done. $\square$

Setting both variables equal gives $a^2 + f(a) \mid af(a) + a$, from which it follows that $f(a) \geq a$. Setting $a=2$ gives $4 + f(2) \mid 2f(2) + 2 \implies 4+f(2) \mid 6$, so $f(2) = 2$. Setting $m=2$ gives $4 + f(n) \mid 4+n$, so $f(n) = n$ for all $n$.

The only solution is $f(n) = n$ for all positive integers $n,$ which clearly works. Let $P(m,n)$ denote the given divisibility relation. $P(2,2)$ gives $4 + f(2) \mid 2 f(2) + 2.$ If $2 f(2) + 2 > f(2) + 4,$ then the divisibility implies $2 f(2) + 2 \ge 2 (f(2) + 4)) = 2 f(2) + 8,$ which is a contradiction. Thus $2 f(2) + 2 = f(2) + 4,$ so $f(2) = 2.$ Next, $P(2,n)$ gives $4 + f(n) \mid 4 + n,$ so $f(n) \le n$ for all positive integers $n.$ Now, let $p$ be an arbitrary prime. $P(p,p)$ gives $p^2 + f(p) \mid p f(p) + p = p(f(p) + 1).$ If $p \nmid p^2 + f(p),$ then $p^2 + f(p) \mid f(p) + 1,$ which is impossible since $p^2 > 1.$ Therefore, $p \mid p^2 + f(p),$ so $p \mid f(p).$ Since $f(p) \le p,$ this implies that $f(p) = p$ for all primes $p.$ Finally, $P(p,n)$ gives $p^2 + f(n) \mid p^2 + n.$ Taking $p$ to be arbitrarily large immediately gives $f(n) = n$ for all positive integers $n,$ as desired.

Let $P(n,m)$ be the assertion. $P(n,f(n))$ gives us $f^{2}(n)+f(n) | f(n)f(f(n))+n $, from this we can easily see that $f(n) | n$. By setting $n=1$, we can get that $f(1)=1$. $P(n,1)$ gives us $1+f(n) | 1+n \Rightarrow f(n) \leq n$ $P(1,n)$ gives us $n^{2}+1 | nf(n)+1 \Rightarrow n^{2} \leq nf(n)$ Then it is easy to conclude that $f(n)=n$ for all positive integers $n$.

Letting $m=f(n)$ gives $(f(n)^2+f(n))|(f(n)f(f(n))+n)$ so $f(n)|n$. Since $f(1)|1$, $f(1)=1$. Plugging in $n=1$ gives $(m^2+1)|(mf(m)+1)$. Consider $x\ge 2$. Since $f(x)<x$ gives $(x^2+1)|(xf(x)+1)<(x^2+1)$ but $f(x)|x$, we must have $\boxed{f(x)=x}$ for all $x$.

I claim the answer is only the identity, it is easy to verify that this works. Take $m = f(n)$, we have $f(n) \mid f(n)^2 + f(n) \mid f(n)f(f(n)) + n$ so $f(n) \mid n$, forcing $f(1) = 1$. Then set $n = 1$, we have $f(1) - 1 \le m (f(m) - m)$, implying $m \ge f(m)$, since we already know $f(m) \mid m$ we have $m = f(m)$ for all $m$ as desired.

Denote the given assertion by $P(m, n)$. We claim the full solution set is $f(x) = x$ for all $x$. This function works, so we now prove it is the only such one. Claim: $f(x) \ge x$ for all $x$. Proof: Observe that$$P(m, 1) \implies m^2 + f(1) \mid mf(m) + 1$$$$\implies mf(m) + 1 \ge m^2 + f(1) \ge m^2 + 1$$$$\implies f(m) \ge m.$$$\square$ Claim: $f$ has a fixed point (in particular, $f(2) = 2$). Proof: Note that $$P(2, 2) \implies 4 + f(2) \mid 2f(2) + 2$$$$ \implies 4 + f(2) \mid 6$$$$\implies 4 + f(2) \le 6 \implies f(2) \le 2.$$But from the first claim, $f(2) \ge 2$. Therefore we indeed have $f(2) = 2$. $\square$ Claim: $f$ is the identity function. Proof: We have that $$P(2, m) \implies 4 + f(n) \mid 4 + n$$$$\implies 4 + f(n) \le 4 + n \implies f(n) \le n.$$But from the first claim $f(n) \ge n$. So we have $f(n) = n$ for all $n$. $\square$ This final claim was what we wanted to prove, QED.

Putting $m=n$ it becomes obvious that $f(m)\geq m$, for any $m$. So, let's define $g:\mathbb Z_{>0} \to \mathbb Z_{\geq 0}$ such that $g(n)=f(n)-n$. Then, $$m^2+g(n)+n\mid m^2+mg(m)+n \implies m^2+g(n)+n \mid mg(m)-g(n) \qquad (1)$$Note that if $g(k)\leq g(1)$ for some $k$ then $k$ must be bounded. So there is a $N$ such that $g(n)\geq g(1)$ for $n\geq N$. Taking $n\geq N$ and putting $m=1$ in $(1)$ we get $$g(n)+n+1\mid g(n)-g(1).$$This forces $g(n)=g(1)$. Now taking $m=n\geq N$ in $(1)$ we get $$n^2+n+g(1)\mid g(1)(n-1).$$For large enough $n$ this forces $g(1)=0$. Thus it follows that $g(n)=0$ for all $n$. Hence, $f(n)=n$ for all $n$ which clearly works.

I thought I posted this before, but maybe not: here is a generalization from Bulgaria's Balkan MO 2014 TST - for a fixed positive integer $k$, consider $m^2 + f(n^k) \mid mf(m) + n^k$. To solve this, note that $m=f(n^k)$ yields that $f(n^k)$ divides $n^k$, in particular $f(1) = 1$. But then $m^2 + 1 \mid mf(m) + 1$, so $f(m) \geq m$ for all $m$, thus $f(n^k) = n^k$ for all $n$. Back to the initial condition, we want $m^2 + n^k \mid mf(m) + n^k$, i.e. $m^2 + n^k \mid m(f(m) - m)$. Fixing $m$, the divisor attains infinitely many values as $n$ varies, thus $m(f(m) - m) = 0$ for all $m$, implying $f(m) = m$ for all $m$, which works.

Put $m=f(n)$ to get $f(n) | n$. Let $f(m)=\frac{m}{d} ;d|m$ and Now $m=n$ to get $m^2 +\frac{m}{d} | \frac{m^2}{d}+m^2$. This implies $dm^2+m |m^2+dm^2$ or $dm+1 |m+d$. Now $dm+1 \leq m+d$ so $d=1$ or $m=1$ so $d=1$ in the latter case as well. Hence $f(m)=m$.

Let $P(m,n)$ denote the given relation. Setting $P(2,2)$ yields that $(f(2)+4)\mid(2f(2)+2)$, so that $(f(2)+4)\mid6$ and thus $f(2)=2$. Now setting $P(2,n)$ yields that $(f(n)+4)\mid(n+4)$, while setting $P(n,2)$ yields that $(n^2+2)\mid(nf(n)+2)$. Since $n$ and $f(n)$ are always positive integers, it follows that $f(n)+4\le n+4$ and $n^2+2\le nf(n)+2$, or equivalently $f(n)=n$, for all integers $n$. Hence $f$ is the identity. $\blacksquare$

uwu~ i think this is different from other ppl Let $P(m,n)$ be the assertion given. $P(1,n)$, yields $1+f(n)\mid f(1)+n$. Let $n=p-f(1)$ for some prime $p$ large enough. Then, $1+f(p-f(1))\mid p$, but this means (as $f\geq 1$ by codomain), $1+f(p-f(1))=p$, or $f(p-f(1))=p-1$. Now, $P(m,p-f(1))$ gives $m^2+p-1\mid mf(m)+p-f(1)$. Hence, $m^2+p-1\mid mf(m)+p-f(1)-m^2-p+1=mf(m)-m^2+1-f(1)$. The RHS is constant for $m$ constant, but the LHS tends to infinity as $p\to \infty$ (there is an infinitude of primes). This is impossible unless the RHS is $0$, or $mf(m)-m^2=f(1)-1$. Factoring, and noting $f(1)\geq 1$, $m(f(m)-m)\geq 0$, hence $f(m)\geq m$. Now, suppose that $f(x)\neq x$ for infinitely many positive integers $x$. Then, there must be arbitrarily large $m$ such that $f(m)>m$, (as equality won't hold) thus by discrete inequality, $f(m)-m\geq 1$ so $f(1)-1=m(f(m)-m)\geq m$ which is a contradiction as the LHS is bounded. Now, this implies that for some constant $C$, $f(x)=x \quad \forall x>C$, now for some $m>C$, considering $P(m,n)$ gives $m^2+f(n)\mid m^2+n$. As the RHS is non-negative, $m^2+f(n)\leq m^2+n$, or $f(n)\leq n$ for all positive integers $n$. Combining with the previous bound of $f(m)\geq m$, we conclude that $f(x)=x$ for all positive integers $x$.

I think I have a pretty unique and short solution. Let $n=m,$ then $m^2+f(m) \leq mf(m)+m \implies (m-f(m))(m-1) \leq 0 \implies f(m) \geq m.$ Now suppose that $n$ is such that $m|f(n).$ Then $m | m^2+f(n) | mf(m)+n \implies m | n.$ Thus letting $m=f(n)$ yields $f(n) | n \implies f(n) \leq n.$ Thus, $f(m)=m$ for all $m,$ and this is the only solution.