too easy problem: My solution: $x=y\Longrightarrow x\mid f(x)$. so there exist a function $g:\mathbb{N}\longrightarrow \mathbb{N}$ such that $f(x)=xg(x)\Longrightarrow x+y\mid xg(x)+yg(y) ,x^2\geq 2g(x)$. notice that $y\equiv -x\pmod{x+y}$ hence $x+y\mid xg(x)-xg(y)$ if $(x,y)=1\Longrightarrow x+y\mid g(x)-g(y) \Longrightarrow$ if $(n,x)=1 , n\mid g(n-x)-g(x) $ now take an arbitrary large odd number $n$ and an arbitrary odd number $x$ such that $(x,n)=1$ then note that: $n\mid 2n\mid g(2n-x)-g(x) (i)$ and $n\mid g(n-x)-g(x) (ii)$ from $(i) , (ii)$ we get $n\mid g(2n-x)-g(n-x)$ Also from $\bigstar$ we get $3n-2x\mid g(2n-x)-g(n-x)$. since $(n,3n-2x)=1$ we get $n(3n-x)\mid g(2n-x)-g(n-x)\Longrightarrow \frac{(2n-x)^2}{2} \geq n(3n-x)\Longleftrightarrow x^2\geq 2n^2$ but the last inquality is false hence $g(2n-x)=g(n-x)$ for every odd $x$ such that $(n,x)=1$ and $2n-x\ge 1395 \bigstar\bigstar$. take two arbitrary large enough integers $a,b\in \mathbb{N}$ such that $(a,b)=1$ and $a\equiv 1\pmod{2}, b\equiv 0\pmod{2}$. take $n=a-b,x=a-2b$ then because $n,x$ are both odd and $(n,x)=1$ from $\bigstar\bigstar$ we get $g(a)=g(b)=t$ hence for a fixed number $b$ there is a constant number $t=g(b)$ such that for infinitely many integers $x$: $t=g(b)=g(x)\Longrightarrow f(x)=tx$. take suffiently large number $x$ such that $f(x)=tx$ then from main problem we get $x+y\mid xt+f(y)$ ($y$ is arbitrary) $\Longrightarrow x+y\mid f(y)-yt$ but the left hand side is larger than the right hand side for large enough $x$ hence we must have $f(y)=yt$ for every $y\in\mathbb{N}$ at last $t\leq \frac{x^2}{2}$ for $x\geq 1395$ hence $t\leq \frac{1395^2}{2}$. so the only solutions are $f(x)=tx$ where $t\leq \frac{1395^2}{2}$ is fixed. Q.E.D

Andria, you lost last condition in the end. $2tx\leq x^3$ for $x\geq 1395$, so $t\leq \frac{1395^2}{2}$ $t \in [1,973012]$

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who is the author of the problem?

sepehrOp wrote: K.N wrote: Find all functions $f: \mathbb N \to \mathbb N$ Such that: 1.for all $x,y\in N$:$x+y|f(x)+f(y)$ 2.for all $x\geq 1395$:$x^3\geq 2f(x)$ we know that for large $x$ we have $a <= \frac{x^2}{2} $ (related to condition 2) so we can see that LHS is increasing and the RHS is going to be little than LHS . so $b=a$. Can you explain this?

math-helli wrote: who is the author of the problem? Mr.jamali

easy problem

Attachments:

Abusing size and primes? Let $P(x,y)$ denote the assertion $x+y|f(x)+f(y)$ Note $P(x,x)$ gives us $x|f(x)$ Now, consider all primes $p$ such that $p>>f(1),f(2)$ Since $f(p)\leq \frac{p^3}{2}$, let $f(p)=ap^2+bp$, where $a,b<p$ $P(p,1)$ gives us $$p+1|f(p)+f(1)=ap^2+bp+f(1)$$$$p+1|(b-a)p+f(1)$$since $p>>f(1)$, this implies $b-a=f(1)$ Similarly, considering $P(p,2)$ gives us $b-2a=f(2)$ Equating the two, we get that $$a=f(1)-f(2)$$$$b=2f(1)-f(2)$$This also implies that for all $p>>f(1),f(2)$, $f(p)=ap^2+bp$ for fixed constants $a,b$ Hence consider two large primes $p,q$ $P(p,q):$ $$p+q|a(p^2+q^2)+b(p+q)$$$$p+q|a(p^2+q^2)$$$$p+q|2apq$$$$p+q|2a$$But note $a\leq \frac{p}{2}$ which implies either $a=0$ or $p+q\leq p$ Hence, $a=0$, which implies that $b=f(1)$ Now, for any arbitrary number $n$, choose a prime $p$ such that $p>>f(n),f(1)$ $P(p,n):$ $$p+n|f(1)p+f(n)$$which implies $f(n)=f(1) \cdot n$ $\forall n$ due to size In order to satisfy the $x^3 \geq 2f(x)$ condition, all functions $f(n)=kn \forall n, \forall k \leq \lfloor \frac{(1395)^2}{2}\rfloor$ work Q.E.D