Snakes wrote: Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f:\mathbb{N}\longrightarrow\mathbb{N}$ such that \[n+f(m)\mid f(n)+nf(m)\]for all $m,n\in \mathbb{N}$ Proposed by Albania https://artofproblemsolving.com/community/c6h1441694p8209574

Note that the condition implies $n+f(m)|f(n)-n^2 \ (*)$. In particular, $1+f(1)|f(1)-1$ so $f(1)=1$. Obviously, $f(n)=n^2$ for all positive integers satisfies the hypothesis, so supppose there is $n_0 \in \mathbb{N}$ such that $f(n_0)\ne n_0^2$. We have that $n_0+f(m)|f(n_0)-n_0^2$, so $f(m)\le n_0+f(m)\le |f(n_0)-n_0^2|$, i.e. $f$ is bounded. Take $m=1$ in the initial relation to get $n+1|f(n)+n$ or equivalently $n+1|f(n)-1$ for all positive integers $n$. However, $f$ is bounded, so we have that $f(n)=1$ for all $n$ big enough, say $n>N$. Take $n>N$ in the hypothesis and note that $n+f(m)|1+nf(m)$ implies $n+f(m)|f(m)^2-1$. Take $n$ large enough to infer that $f(m)=1$ for all positive integers $m$, which indeed is a solution of the problem.

Note that $f(x)=x^2$ does satisfies the conditions of the problem. Now suppose that $f\not\equiv\text{id}^2$.Then there must be $n\in\mathbb{N}$ such that $|f(n)-n^2|\neq 0.$ Then $n+f(m)|f(n)+nf(m)\forall m\in\mathbb{N}$ is easily seen to imply that $n+f(m)| f(n)-n^2 \forall m\in\mathbb{N}.$ Thus $1+f(m)\le n+f(m)\le |n^2-f(m)|$ and so $f(m)\le C$ for some constant $C$(we have $C=|f(n)-n^2|-1$). Fix some random number $m$ and let $c=f(m)$.Then $n+c|f(n)+cn~\forall n\in\mathbb{N}$ and so $n+c|f(n)-c^2~\forall n\in\mathbb{N}.$ Therefore,for $n>C+c^2$,if $f(n)\neq c^2$ we have $$C+c^2<n<n+c\le |f(n)-c^2|\le C+c^2$$contradiction! Thus $f(n)=c^2$ for big enough $n$.So if we let $x,y$ be two arbitrary positive integers,by the above we have that $f(n)=f(x)^2$ for big enough $n$ and $f(n)=f(y)^2$ for big enough $n$.Hence $f(n)=f(x)^2=f(y)^2$ for big enough $n$ and it follows that $f(x)=f(y)$ i.e. $f$ is constant.Now it's rather easy to prove that we must have $f\equiv 1$.

This problem was proposed by me.

My solution: http://www.mathematica.gr/forum/viewtopic.php?p=282399#p282399

First note that $1+f(1)|2f(1)$ so $f(1)=1$. Suppose that there exists an integer $k$ such that $f(k) \neq k^2$. By the condition we have $k+f(m)|k^2-f(k)$ so $f$ is bounded. Furthermore we know that $n+1|f(n)+n$ from $m=1$ in the condition so $f(n) \equiv 1 \pmod{n+1}$ but since $f$ is bounded we have $f(n) = 1$ for $n$ sufficiently large. To finish note that $n+f(m)|nf(m)+f(m)^2$ so $n+f(m)|f(m)^2-f(n)$. Take a very large $n$ in the previous relationship to get $n+f(m)|f(m)^2-1$ thus $f(m) = 1$ for all $m \in \mathbb{N}$ and we get the solution $f(n) \equiv 1$. If such a $k$ doesn't exist we get the solution $f(n) \equiv n^2$.

Let $P(n,m)$ the assertion of $n+f(m)|f(n)+nf(m)$ $P(n,n):$ $n+f(n)|(n+1)f(n)$ $\Longrightarrow$ $n+f(n)|(n+1)f(n)-n(f(n)+n)$ $\Longrightarrow$ $n+f(n)|n^2-f(n)$, if exist $n\in \mathbb{N}$ such that $f(n)>n^2$ $\Longrightarrow$ $n+f(n)|f(n)-n^2$, but $n+f(n)>f(n)-n^2$, hence is a contradicction $\Longrightarrow$ $n^2\geq f(n)$ for all $n\in \mathbb{N}$ $P(1,1):$ $1+f(1)|2f(1)$ $\Longrightarrow$ $1+f(1)|2(1+f(1))-2f(1)$ $\Longrightarrow$ $1+f(1)|2$ $\Longrightarrow$ $f(1)=1$ $P(2,2):$ $2+f(2)|3f(2)$ $\Longrightarrow$ $2+f(2)|3(2+f(2))-3f(2)$ $\Longrightarrow$ $2+f(2)|6$ $\Longrightarrow$ $f(2)=1$ or $f(2)=4$ If $f(2)=1:$ $P(2,m):$ $2+f(m)|1+2f(m)$ $\Longrightarrow$ $2+f(m)|2(2+f(m))-1-2f(m)$ $\Longrightarrow$ $2+f(m)|3$ $\Longrightarrow$ $f(m)=1$ for all $m\in \mathbb{N}$ which is solution. If $f(2)=4:$ $P(3k,2):$ $3k+4|f(3k)+12k$ $\Longrightarrow$ $3k+4|f(3k)+12k-3k(3k+4)$ $\Longrightarrow$ $3k+4|9k^2-f(3k)...(1)$ $P(3k,1):$ $3k+1|f(3k)+3k$ $\Longrightarrow$ $3k+1|f(3k)+3k-3k(3k+1)$ $\Longrightarrow$ $3k+1|9k^2-f(3k)...(2)$ $\Longrightarrow$ by $(1)$ and $(2)$ we get $(3k+1)(3k+4)|9k^2-f(3k)$, so from $(3k+1)(3k+4)>9k^2-f(3k)\geq 0$ we get $f(3k)=9k^2$ for all $k\in \mathbb{N}$ On the other hand, let $q$ be a number such that $9q^2>n^2-f(n)$, by $P(n,3q):$ $n+9q^2|f(n)+9q^2n$ $\Longrightarrow$ $n+9q^2|n(n+9q^2)-f(n)-9q^2n$ $\Longrightarrow$ $n+9q^2|n^2-f(n)$, so from $n+9q^2>n^2-f(n)\geq 0$ we get $f(n)=n^2$, hence $f(n)=n^2$ for all $n\in \mathbb{N}$ Hence all functions $f$ are: $f(n)=1$ and $f(n)=n^2$.

Setting $m=n$ gives: $n+f(n) \vert f(n)+n f(n) \implies n+f(n) \vert f(n)-n^2 \implies n+f(n) \vert n+n^2$ As both sides are positive we have: $n+f(n) \leq n+n^2 \implies f(n) \leq n^2$ Clearly this gives $f(1) \leq 1 \implies f(1)=1$. We also get $f(2) \leq 4$. Setting $n=2,m=1$ gives: $2+f(1) \vert f(2)+2f(1) \implies 3 \vert f(2)-1$ So the only possible values for $f(2)$ are $1,4$. It $f(2)=4$ then we proceed by induction on $n \geq 3$. Assume $f(k)=k^2$ for all $k \leq n-1$. Setting $m=k$ we see: $n+f(k) \vert f(n)+n f(k) \implies n+k^2 \vert f(n)+nk^2 \implies n+k^2 \vert f(n)-n^2$. If $f(n)=n^2$ we're done otherwise we have $f(n)<n^2$ so $n+k^2 \vert n^2-f(n)$. Setting $k=n-1$ we see: $n^2-n+1 \vert n^2-f(n)$ but clearly as $n^2-n+1>\dfrac{n^2}{2}>\dfrac{n^2-f(n)}{2} \Leftrightarrow (n-1)^2>0$ which is obviously true and $n^2-f(n)>0$ we have: $n^2-n+1=n^2-f(n) \implies f(n)=n-1$ But setting $k=n-2$ gives: $n^2-3n+4 \vert n^2-f(n)=n^2-n+1$ But $n^2-3n+4<n^2-n+1<2(n^2-3n+4)$ as the $RHS \Leftrightarrow (n-3)^2+(n-2)>0$ which is obvious for $n \geq 3$ yielding a contradiction so $f(n)=n^2$ and hence we are finished by induction. If $f(2)=1$ then setting $n=2$ gives: $2+f(m) \vert f(2)+2f(m)=1+2f(m)$ But as $2+f(m) \leq 1+2f(m)$ (as $f(m) \geq 1$) and $1+2f(m)<2(2+f(m))$ we must have: $2+f(m)=1+2f(m) \implies f(m)=1$. We now have the two solutions: $f(n)=n^2$ or $f(n)=1$.

Let $P(m,n)$ the given relation.. Step 1: The relation $P(1,1)$ gives that $1+f(1)\mid 2f(1)$, so $f(1)=1$. Step 2: The relation $P(n,n)$ gives us $n+f(n)\mid (n+1)f(n)$, so $n+f(n)\mid n(n+1)$, and thus it follows that $n+f(n)\leq n(n+1)$ or $f(n)\leq n^2$ for all positive integers $n$. Step 3: The relation $P(p,p)$ with $p$ being a prime gives $p+f(p)\mid (p+1)f(p)$. If $p\nmid f(p)$ then $(f(p),p+f(p))=1$ so $p+f(p)\mid p+1$, which gives $f(p)\leq 1$, so $f(p)=1$. This means that $f(p)=1$, or $p\mid f(p)$. We write $A=\{p\in\mathbb{P},\ p\mid f(p)\}$ and $B=\{p\in\mathbb{P},\ f(p)=1\}$. Step 4: From Step 3 we have that at least one of $A$ and $B$ is an infinite set. We have two cases: Case α) If $A$ is infinite, let a positive integer $n$. From $P(n,p)$ with $p\in A$ we have that $n+f(p)\mid n^2-f(n)$, so $n+f(p)\leq n^2-f(n)$ and since $p\in A$, if $f(n)\neq n^2$, we will have $p\leq f(p)\leq n^2-n-f(n)$, which is a contradiction since $A$ is infinite. It follows that $f(n)=n^2$. Case β) If $B$ is infinite, let $n$ be a positive integer. Then $P(p,n)$ with $p\in B$ gives $p+f(n)\mid p^2-f(p)$ and since $p\in B$ we will have that $p+f(n)\mid p^2-1$. Adding and subtracting $f^2(n)$ we get that $p+f(n)\mid f^2(n)-1$. So, $p\leq f^2(n)-f(n)-1$, if $f^2(n)\neq 1$, which is a contradiction since $B$ is infinite. This means that $f(n)=1$ for all $n$.

Let $P(n,m)$ be the given relation. Then $P(1,1)$ implies, \[1+f(1) \mid 2f(1)\]So, $f(1)=1$ \[P(n,n) \implies n+f(n) \mid f(n)+nf(n)\]\[ \implies n+f(n) \mid (n+1)(n+f(n))-(f(n)+nf(n))=n^2+n\] From this, we have $f(n) \leq n^2$ And, $2+f(2) \mid 6 \implies f(2)=1$ or $4$ If f(2)=1, \[P(2,m) \implies 2+f(m) \mid 1+2f(m)\]\[ \implies 2+f(m) \mid 2(2+f(m)-(1+2f(m))=3\]So, $f(n)=1$ $\forall n\in\mathbb{N}$ Now consider the case for $f(2)=4$. We will prove by induction that $f(n)=n^2$. The base case is done for $n=1$ and $n=2$. Now for inductive hypothesis, assume that $f(n)=n^2$ $\forall 1\leq n\leq k$. Then, \[P(k+1,k)\implies k+1+k^2 \mid f(k+1)+k^3+k^2\]\[\implies k^2+k+1 \mid f(k+1)-k\] Definitely it's not possible to have $1\leq f(k+1)<k$. Also $f(k+1)\neq k$ as then by taking $P(k+1,1)$, we have \[k+2\mid 2k+1 \implies k+2 \mid 3\]So, $k=1$ which is a contradiction as we have $f(2)=4$. So, \[k^2+k+1 \leq f(k+1)-k\]\[\implies (k+1)^2\leq f(k+1)\] But, we also have $f(n)\leq n^2$ $\forall n\in\mathbb{N}$. So, $f(k+1)=(k+1)^2$ and we are done by induction. It's not hard to verify that both $f(n)=1$ and $f(n)=n^2$ satisfies the given equation.

$f(n)=1$

My tears of Today Here's my solution!! Sorry if there is any error. In my solution I didn't type choice 3. Choice 3: Replace $ n=f (t) $..... in 25^{th} line of my solution. AND THEN THE REST OF MY SOLUTION will be continued. Because choice 3 is out the $f (x)$ from choice of it's specificity that we cannot garuntee it's a polynomial function or not.

@Safal_db I think your solution is flawed, since you seem to be assuming $f$ is a polynomial, which is not given.

Ankoganit wrote: @Safal_db I think your solution is flawed, since you seem to be assuming $f$ is a polynomial, which is not given. Note the place in my solution where I write $n=f (t)$ from that part to rest of the solution my solution is not about f being in polynomial the catch I not mention in my solution that if $n+f(m)$ divides $f (n)-n^{2} $ then $f (m)$ must be a dependable on some function of $k (n)$ otherwise for any independent value of $m$ , $f (n)=n^{2}$ must hold as $n+f (m) $ divide $f (n)-n^{2}=0$ there I don't assume that f is a polynominal function but although I don't assume that but still the answer will be a polynomial function after my calculation. I Mention cases in my analysis But Forget to Write Choice 3. That is $n=f (t) $ and choice 3 is not a polynomial function. Am I right? @Ankoganit

@Ankoganitganit I got my mistake Thank you. Yes in Choice 3 I need to describe it more explicitly!!! I am giving a new solution of this question tommorow

Snakes wrote: Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f:\mathbb{N}\longrightarrow\mathbb{N}$ such that \[n+f(m)\mid f(n)+nf(m)\]for all $m,n\in \mathbb{N}$ Proposed by Dorlir Ahmeti, Albania A little bit different but basically similar: My soloution:(hope it is corret!) Put $m=n=1$ we get $f(1)+1|2$ so $f(1)=1$ Now put $m=1$ we get : $n+1|f(n)+n$ so $n+1|f(n)-1$ for all $n$ Now let $g(n)$ be a function such that $f(n)-1=(n+1)g(n)$ for all $n$ By Putting this we get that: $n+(m+1)g(m)+1|(n+1)g(n)+1+n((m+1)g(m)+1)$ And we know that: $(m+1)g(m)+1 \equiv -n$ (mod $n+(m+1)g(m)+1$) So we get that : $n+(m+1)g(m)+1|(n+1)g(n)+1-n^2$(*) Now fix $n$ and give large $m$ in (*): We get two cases : 1) $g(n)=0$ that give us the solution $f(n)=1$ 2) $(n+1)g(n)+1-n^2=0$ so we get that $g(n)=n-1$ so we get the solution: $f(n)=n^2$

is this solution valid? f(1)=1 and(call this first) n+f(n) divides n(n+1) implying f(n)<=n^2 and(this one second) 1+n divides f(n)-1 . now we put n=n-1 in the last and get n divides f(n-1)-1 from which we get that f(n-1) is congruent 1 mod n and from first f(n-1)<=(n-1)^2 and f(n-1) can be 1 ; n+1 ; (n-1)^2 . check all cases easly and see that sols are f(n)=1 and f(n)=n^2

The only solutions are $\boxed{f\equiv 1}$ and $\boxed{f(n)=n^2}$, which work. Let $P(m,n)$ denote the given assertion. The problem condition implies that \[n+f(m)\mid (n^2+nf(m))-(f(n)+nf(m)) = n^2-f(n)\] If $f$ is not upper bounded then taking $m$ such that $f(m)$ is sufficiently large gives $f(n)=n^2$ for each positive integer $n$. Now suppose $f$ is upper bounded. $P(1,1): 1+f(1)\mid 2f(1)\implies f(1)=1$. $P(n,1): n+1\mid f(n)+n$, so $n+1\mid (f(n)+n)-(n+1) = f(n)-1$. Thus, either $f(n)=1$ or $f(n)\ge n+2$. If $f(n)\ne 1$ for infinitely many integers, then we get that $f$ is not upper bounded. So for all sufficiently large positive integers $n$, $f(n)=1$. Note that if $f(n)=1$, for each positive integer $m$, we have \[n+f(m)\mid nf(m)+1\implies n+f(m)\mid (nf(m)+f(m)^2) - (nf(m)+1) = f(m)^2 -1 \] Taking $n$ sufficiently large gives $f(m)=1$ for each positive integer $m$.

$n+f(m) | f(n) + nf(m) \implies n + f(m) | f(n) - n^2$. obviously $f(n) = n^2$ works so assume that there exists $k$ such that $f(k) \neq k^2$. Now let $n = k$ and vary $m$ since $n + f(m) | f(n) - n^2$ so $f(m)$ is bounded. $P(1,1) : 1+f(1) | 2f(1)$ and since $2(1+f(1)) > 2f(1)$ we have $f(1) + 1 = 2f(1) \implies f(1) = 1$. $P(n,1) : n+1 | f(n) + n \implies n+1 | f(n) - 1 \implies f(n) \ge n+2$ or $f(n) = 1$ which means for $n > N$ we have $f(n) = 1$ for a fixed $N$. $P(n>N,m) : n + f(m) | 1 + nf(m) \implies n+f(m) | f(m)^2 - 1$ so for large enough $n$ we have $f(m)^2 - 1 = 0 \implies f(m) = 1$. Answers $: f(n) = n^2 , f(n) = 1$.

Like the above solutions, we can easily obtain that: $f(1)=1$ and $f(2) \in \{1,4 \}$ $Case \ 1:f(2)=1$ Set $n=2$ to get $2+f(m) | 1+2f(m) \ \forall m \in \mathbb{N} $ or $2+f(m) | 3 \ \forall m \in \mathbb{N} $ Thus $ f(m) =1 \ \forall m \in \mathbb{N} (1) $ $Case \ 2:f(2)=4$ We'll prove that $f(n) > 1 \ \forall n\geq 12$ Assume that $f(a)=1$ for some $a \geq 12$ Set $n=a, m=2$ to get: $a+4 | 1+4a $ or $n+4 | 15$ which is cleary false for $a \geq 12$ so such $a$ doesn't exists. Set $m=1, n\geq 12$ $n+1 | f(n) + n $ $\leftrightarrow n+1 | f(n) -1$ $\rightarrow f(n) \geq n+2 \ \forall n \geq 12$ Now take m big enough such that $f(m)> |f(n)-n^{2}|$ We get: $n+f(m) | f(n)+nf(m) $ $\leftrightarrow n+f(m)|f(n)-n^2 \ \forall n \in \mathbb{N}$ Which gives $f(n)=n^{2} \ \forall m \in \mathbb{N} (2)$ It's easy to check that both $(1)$ and $(2)$ satisfy and we're done.

The answer is $f(x)=x^2$ and $f(x)=1$ $f(m)+n|f(n)+nf(m) \Rightarrow f(m)+n|f(n)+n(-n)$ so if the function is not bounded then for any $n$ : $f(n)-n^2=0 \Rightarrow f(n)=n^2$ but we know : $f(1)+1|f(1)-1 \Rightarrow f(1)=1$ If we put $m=1: n+1|f(n)+n \Rightarrow n+1|f(n)-1$ It means $f(n)=1$ or $f(n)\ge n+2$ and for any $m>N:f(m)=1$ because if infinite times $f(m)>m$ it concludes $f(n)=n^2$ so we put a large $t$ for some $m$ such that $tf(m)\ge N$ then $f(m)+tf(m)| tf(m)^2+1$ it means for any $m: f(m)=1$

The solutions are $\boxed{f(n)=1}$ and $\boxed{f(n)=n^2}$ Let $P(n,m)$ be the assertion \[n+f(m)\mid f(n)+nf(m)\]$P(1,1)$ $\implies$ $f(1)=1$ $P(n,1)$ $\implies$ $n+1 \mid f(n)-1$ define $k_n$ as $f(n)=nk_n+k_n+1$ $(1)$ $P(n,n)$ $\implies$ $n+f(n) \mid n(n+1)$ $\implies$ $k_n+1 \mid n$ $(2)$ $p$ is a prime number $n=p$ on $(2)$ we can see $k_p=1,p-1$ so $f(p)=p^2, 1$ for prime $p$ for $f(p)=p^2$ $P(n,p)$ $\implies$ $p^2+n \mid n^2-f(n)$ from prime numbers infinite $\boxed{f(n)=n^2}$ for $f(p)=1$ $P(p,n)$ $\implies$ $p+f(m) \mid f(m)^2-1$ like above $\boxed{f(n)=1}$

The only solutions are $\boxed{f(n)=1\forall n\in\mathbb{N}}$ and $\boxed{f(n)=n^2\forall n\in\mathbb{N}}$ Note that $f(n)+nf(m)\equiv0\pmod{n+f(m)}\Longrightarrow f(n)-n^2\equiv0\pmod{n+f(m)}\Longrightarrow n+f(m)\mid f(n)-n^2$ We show that $f(n)\leq n^2$. Suppose $f(n)>n^2\Longrightarrow n+f(m)\leq f(n)-n^2\Longrightarrow f(n)\geq n^2+n+f(m)$. Suppose there is another $k$ for which $f(k)>k^2$. Then, $f(k)\geq k^2+k+f(n)$, while $f(n)\geq n^2+n+f(k)$. Adding these two equations gives contradiction. So, there is at most one $f(n)$ for which $f(n)>n^2$. Call this $n_0$. Case 1) $f(n)\leq n^2\forall n\in\mathbb{N}$ (This is the case where '$n_0$' does not exist.) Clearly $f(1)=1$. Suppose $f(n)<n^2$ (taken $n\ne1$). Then, $n^2-f(n)\geq n+f(m)\Longrightarrow f(m)+f(n)\leq n^2-n\forall m\in\mathbb{N}$. Let $n_1$ be the minimum possible $n$ for which $f(n_1)<n_1^2$. Then, $f(m)+f(n_1)<n_1^2-n_1\Longrightarrow f(m)<n_1^2-n_1$. It follows that for $m\geq n_1$, $f(m)<m^2$. If $n_1=2$, then we get $f(m)+f(2)\leq2\Longrightarrow f(m)=f(2)=1$, or $f\equiv1$. (This is a valid solution as it satisfies the original equation.) Now suppose that $f(n)$ is not a constant function. (The only constant function that satisfies is $f(n)=1$.) Then, $f(2)=2^2=4$ and $n_1>2$. It is also true that for $n<n_1$, $f(n)=n^2$. We have $n_1+f(n_1-1)\mid n_1^2-f(n_1)\Longrightarrow n_1^2-n_1+1\mid n_1^2-f(n_1)$. Since $2(n_1^2-n_1+1)>n_1^2>n_1^2-f(n_1)$, it follows that $n_1^2-n_1+1=n_1^2-f(n_1)\Longrightarrow f(n_1)=n_1-1$. But $n_1+1=n_1+f(1)\mid n_1^2-f(n_1)=n_1^2-n_1+1\Longrightarrow n_1+1\mid3\Longrightarrow n_1=2$, a contradiction. This means that either $f(n)=1\forall n\in\mathbb{N}$ or $f(n)=n^2\forall n\in\mathbb{N}$ (in this case). Case 2) There exists $n_0$ for which $f(n_0)>n_0^2$. Since $n+f(m)\mid f(n)-n^2$, $f(1)+1\mid f(1)-1\Longrightarrow f(1)+1\mid2\Longrightarrow f(1)=1$. So, $n_0>1$. Note that the proof given in Case 1 works for all numbers except for $n_0$. We have $2+f(n_0)\mid f(2)-4$. If $f(2)=1$ (and subsequently $f(n)=1\forall n\in\mathbb{N}, n\ne n_0$), we get $2+f(n_0)\mid3\Longrightarrow f(n_0)=1$, contradicting with $f(n_0)\geq n_0^2+n_0+f(m)=n_0^2+n_0+1$. For $f(n)=n^2\forall n\ne n_0$, we have $f(n_0)\geq n_0^2+n_0+f(m)=n_0^2+n_0+m^2$. But if $m$ is taken arbitrarily large, $f(n_0)$ blows up to infinity, contradicting the fact that $f(n_0)\in\mathbb{N}$. Therefore the only solutions are the aforementioned. $\blacksquare$

Let $P(n,m)$ be the assertation $n+f(m)|f(n)+nf(m)$ $P(1,1) \Rightarrow 1+f(1)|2f(1) \Rightarrow f(1)=1$ Let $p$ be an arbitrary prime number: $P(p,p) \Rightarrow p+f(p)|f(p)+pf(p) \Rightarrow p+f(p)|p^2+p$ $P(p,1) \Rightarrow p+1|f(p)+p$ If $p|f(p)$: $g(p)=\frac{f(p)}{p}$ $\Rightarrow \begin{cases} P(p,p) \Rightarrow g(p)+1|p+1 \\ P(p,1) \Rightarrow p+1|g(p)+1 \end{cases}$ $\Rightarrow$ $g(p)=p \Rightarrow f(p)=p^2$ If $p\nmid f(p)$: $P(p,p) \Rightarrow p+f(p)|p+1 \Rightarrow f(p)=1$ We got that for any prime $f(p)=1$ or $f(p)=p^2$ Let's write all primes numbers into: $A=\{p\in\mathbb{P}, f(p)=1\}$, $B=\{p\in\mathbb{P},f(p)=p^2\}$ Case 1: $A$ is empty $\Rightarrow$ $p\in\mathbb{P},f(p)=p^2$ $\Rightarrow$ $P(n,p)$ $\Rightarrow$ $n+p^2|f(n)+p^2n \Rightarrow n+p^2|f(n)-n^2$ Let $p>|f(n)-n^2|$ we get that $f(n)-n^2=0 \Rightarrow$ $\boxed{f(n)=n^2}$ Case 2: $A$ is non-empty: Let $a$ be the smallest prime in $A$. If there exists, let $b$ be an arbitrary prime in $B$. $P(a,b) \Rightarrow a+b^2|1+ab^2 \Rightarrow a+b^2|a^2-1 \Rightarrow$ $a+b^2\leq a^2-1 \Rightarrow b<a \Rightarrow$ $B$ is finite, $A$ in inifinite; Let $p \in A, p>f(n)^2 \Rightarrow P(p,n) \Rightarrow p+f(n)|1+pf(n) \Rightarrow$ $p+f(n)|p^2-1, p+f(n)|(p+f(n))(p-f(n)) \Rightarrow$ $p+f(n)|f(n)^2-1 \Rightarrow f(n)^2-1=0 \Rightarrow$ $\boxed{f(n)=1}$

$P(1,1)$ $f(1)+1 | 2f(1)$ $f(1)+1 | f(1)-1$ so $f(1)=1$ Let $p$ be an arbitrary prime number $P(p,p)$ $p+f(p) | f(p)+pf(p)$ 1.Case: $\gcd(p,f(p))=1$ and call these prime numbers at set $Q$ then $p+f(p) | p+1$ so $f(p)=1$ 2.Case: $\gcd(p,f(p))=p$ and call these prime numbers at set $R$ 1.Assume set $R$ is infinite. then $P(n,p)$ $n+f(p) | f(n)+nf(p)$ $n+f(p) | f(n) - n^2 $ if $f(n) \ne n^2$ then $f(n)-n^2-n \geq f(p) $ take $p$ sufficiently large gives us $f(n)=n^2$ 2.Assume set $Q$ is infinite $P(p,m)$ $p+f(m) | 1+ pf(m)$ $p+f(m) |f(m)^2-1$ Let's take $p \geq f(m)^2-f(m)$ then we get contradicition iff $f(m) \ne 1$ if $f(m)=1$ the equation fits very well so that only solutions are $f(n)=n^2$ and $f(n)=1$

Let $P(n,m)$ be the assertion. From $P(1,1)$ we have $f(1)=1.$ Then from $P(2,2)$ we get $f(2)+2\mid 3f(2).$ Since $f(2)+2\mid 3f(2)+6$ we have $f(2)+2\mid 6 \implies f(2)\in\{1,4\}.$ $\textcolor{red}{Case 1:} f(2)=1.$ From $P(2,m)$ we have $f(m)+2\mid 2f(m)+1$ but, $f(m)+2\mid 2f(m)+4\implies f(m)+2\mid 3$ meaning in this case $f\equiv 1$ is the only solution. $\textcolor{red}{Case 2:} f(2)=4.$ Let's prove from induction $f(x)=x^{2}.$ Say $f(x)=x^{2}$ for all integers $1\leq x<x+1.$ From $P(x+1,x)$ we have $$f(x+1)\geq(x+1)^{2}.$$Since $n+f(m)\mid nf(m)+f(n)$ and $n+f(m)\mid n^{2}+nf(m)$ we have $$n+f(m)\mid f(n)-n^2.$$For $m=n$ we got $f(n)+n\leq\left|f(n)-n^{2}\right|$ meaning $f(n)\leq n^{2}.$ So we have $$\left(x+1\right)^{2}\geq f(x+1)\geq \left(x+1\right)^{2}.$$Thus $f(x+1)=x+1$. So the solutions are $\boxed{f\equiv 1, f(x)=x^{2}}$

I think a new solution? First, assume $f$ is bounded. Let $f(a) = c$ be the largest possible value of $f$. $P(f(a), a)$ gives $2c \mid f(c) + f(c)^2 \implies c \mid f(c) \implies c \leq f(c)$. If $f(c) \neq c$, then $f(c) > c$, a contradiction, so $f(c) = c$. $P(c, n)$ gives $n + c \mid f(n) + nc \implies n + c \mid c^2 - f(n)$. Take $n$ infinitely large, which gives $n + c > c^2 - f(n) \implies c^2 - f(n) = 0 \implies f(n) = c^2$. If $c \neq 1$, $c^2 > c$, a contradiction, so $c = 1$, which means $f \equiv 1$, which clearly works. If $f$ is unbounded, we have $n + f(m) \mid f(n) + nf(m) - n^2 - nf(m) \implies n + f(m) \mid f(n) - n^2$. Take $f(m)$ to be infinitely large, which must yield $f(n) = n^2$ for all $n$, so we are done.

i hope my solution is valid This solution is basically a little bit of variation from @HiroHamada's solution, just to limit on the cases. Clearly $\boxed{f(n)=1\quad\forall n\in\mathbb{N}}$ is a solution, so we'll look for nonzero solutions. Let $P(m,n)$ denote the assertion of the given function. Set $x=y=1$ so we can have $f(1)+1\mid 2f(1)\Longleftrightarrow f(1)+1\mid 2$ so $f(1)=1$. Now take $P(n,n)$ to get \[f(n)+n\mid (n+1)f(n)\]\[f(n)+n\mid nf(n)-n\]\[f(n)+n\mid n^2+n\]so we can have $f(n)\le n^2$. Next, taking $P(1,n)$ yields $n+1\mid f(n)+n\Longleftrightarrow n+1\mid f(n)-1\dots(\star)$. Since that $f(n)\ne 1$ for all $n\ne 1$, we may have $n+1\le f(n)-1\Longleftrightarrow f(n)\ge n+2$. So we conclude that $n+2\le f(n)\le n^2\dots(\circ)$. Substitute $n=2$, so we have $4\le f(2)\le 4\Longleftrightarrow f(2)=4$. We see that $f(1)=1=1^2$ and $f(2)=4=2^2$. Now use induction over $\mathbb{N}$, assume that $f(k)=k^2$ for some $n=k\in\mathbb{Z}$. It suffices to prove that $f(k+1)=k+1$ for $n=k+1$. Take $P(k,k+1)$, so we have \[k^2+k+1\mid f(k+1)+k^3+k^2\]\[k^2+k+1\mid f(k+1)-k\Longrightarrow k^2+k+1\le |f(k+1)-k|\]Now if $f(k+1)-k<0\Longrightarrow 1\le f(k+1)<k$, we have $k^2+k+1\le k-f(k+1)\Longleftrightarrow f(k+1)\le -k^2-1\le -2<1$, a contradiction. If $f(k+1)-k=0\Longleftrightarrow f(k+1)=k$, setting $k+1\rightarrow k$ yields $f(k)=k-1$, so from $(\star)$ we may have $k+1\mid f(k)-1\Longleftrightarrow k+1\mid k-2$, which is impossible. So we need to have $f(k+1)-k>0$ i.e. \[k^2+k+1\le f(k+1)-k\]\[f(k+1)\ge k^2+2k+1=(k+1)^2\]But from $(\circ)$ we know that $f(k+1)\le (k+1)^2$ and so $f(k+1)=(k+1)^2$, finishing our induction. In conclusion, the only functions that satisfy the problem are $\boxed{f(n)=1\quad\forall n\in\mathbb{N}}$ and $\boxed{f(n)=n^2\quad\forall n\in\mathbb{N}}$, which clearly work. Therefore, we're done. $\blacksquare$

THIS SOLUTION IS WAY DIFFERENT THAN OTHERS

PART 2 OF THE SOLUTION ABOVE

PART3 OF THE SOLUTION

Solution by Mhremath and Sadigly $P(1;1)\Rightarrow 1+f(1)\mid 2f(1)\hspace{10mm} f(1)=1$ $P(2;2)\Rightarrow 2+f(2)\mid 3f(2)$. Either $f(2)=1$ or $f(2)=4$ $P(n;1)\Rightarrow n+1\mid f(n)+n\Rightarrow n+1\mid f(n)-1$ This could mean 3 things. Case 1) $\forall x\in\mathbb{N} \hspace{2mm} f(x)=1$ Case 2) $\forall x\in\mathbb{N} \hspace{2mm} f(x)-1\geq x+1$ Case 3) $f$ is a weird function that sometimes take $1$, while sometimes is greater(or equal) than $n+2$ Case 1. This case is trivial. Indeed,$f(x)=1$ is actually an answer Case 2. We have $n+f(m)\mid f(n)+nf(m)-(n^2-nf(m))\Rightarrow n+f(m)\mid f(n)-n^2$. So,as $m\rightarrow\infty\Rightarrow n+f(m)\rightarrow\infty$. We can say that $f(n)-n^2=0\Rightarrow f(n)=n^2$ Case 3. Plugging $P(n;2)$ into $n+f(m)\mid f(n)+nf(m)$ gives either $n+4\mid f(n)+4n$ or $n+1\mid f(n)+n$, depending on value of $f(2)$ Subcase 1. $f(2)=4\Rightarrow n+4\mid f(n)+4n-(4n+16)$. We have $n+4\mid f(n)-16$. If $f(n)=1$ for some random value $a\in\mathbb{N}$, we must have $a+4\mid-15$. $$a\in\{1;11\}$$ $f(11)\neq1$, because $P(11;2)$ gives contradiction. Notice that this subcase became equivilaent with Case 2. . Subcase 2. $f(2)=1$. Plugging $P(2;m)$ gives $2+f(m)\mid1+2f(m)-(4+2f(m))\Rightarrow 2+f(m)\mid3$. We have $\forall n\in\mathbb{N}\hspace{2mm}f(n)=1$ So,only functions that satisfies our condition is $f\equiv1$ and $f(n)=n^2$