Nice problem here is my solution(i am new so my latex might be erroneous) Lemma 1, $f$ is injective. Fairly simple. Put $m=1$ in the relation to get $f(f(n))=n+f(2015)$. Then $f(x)=f(y)$ gives $x+f(2015)=y+f(2015)$ So $x=y$. Lemma 2 $f(f(n))-f(f(k))=n-k$ for all $k,n$ in the set of positive integers. Obvious. Lemma 3 $f(mf(n))-f(mf(k))=n-k$ Again obvious. Now by Lemma 2 and 3 we have $f(mf(n))-f(n)=c$ where c is some constant. Clearly $c=f(2015m)-f(2015)$. So $f(2015m)=a$ where $a$ is a constant. Put this in the original equation. So now we have $f(mf(n))=n+f(2015m)=n+f(2015(m+1))=f((m+1)f(n))$. The injectivity then gives $mf(n)=(m+1)f(n)$. So $f(n)=0$ for all n. But 0 is out of $f$ range so no solution.

How does $f(mf(n)) - f(n) = c$ follow from lemma 2 and 3?

Khazix wrote: How does $f(mf(n)) - f(n) = c$ follow from lemma 2 and 3? We have $f(f(n))-f(f(k))=n-k=f(mf(n))-f(mf(k)),\forall m,n\in\mathbb{Z_+}$,so $f(mf(n))-f(n)=f(mf(k))-f(k),\forall m,n\in\mathbb{Z_+}$.From here it follows that the number $f(mf(n))-f(n)$ is constant.

huricane wrote: We have $f(f(n))-f(f(k))=n-k=f(mf(n))-f(mf(k)),\forall m,n\in\mathbb{Z_+}$,so $f(mf(n))-f(n)=f(mf(k))-f(k),\forall m,n\in\mathbb{Z_+}$.From here it follows that the number $f(mf(n))-f(n)$ is constant. From $f(f(n)) - f(f(k)) = f(mf(n)) - f(mf(k))$ it seems like all we can deduce is that $f(mf(n)) - f(f(n)) = f(mf(k)) - f(f(k))$. I don't see how we can get $f(mf(n)) - f(n) = f(mf(k)) - f(k)$.

@khazix Well i probably am not sure about what exactly your query is but hope this might resolve. Subtract the two equations from Lemma 2 and 3 to get for all $m,n,k$ in $N$, $f(mf(n))-f(n)=f(mf(k))-f(k)$. From here we can see that the expression is constant. Still i will recheck my proof and convey it to you.

Tsarik wrote: Find all functions $f : \mathbb{Z}_{+} \rightarrow \mathbb{Z}_{+}$ that satisfy $f(mf(n)) = n+f(2015m)$ for all $m,n \in \mathbb{Z}_{+}$. My solution: Let $P(m,n)$be the assertion that "$f(mf(n))=n+f(2015m)$" for all $m,n \in \mathbb{Z}_{+}$. (1) $f(f(n))=n+f(2015)$ for all $n \in \mathbb{Z}_{+}$. proof: consider $P(1,n)$ (2) $f$ is one-to-one function on $\mathbb{Z}_{+}$.[obviously from (1)] (3) $f(f(f(4030)))=f(4030)+f(2015)$ proof: consider $P(1,f(4030))$ (4) $f(2f(f(2015)))=f(2015)+f(4030)$ proof: consider $P(2,f(2015))$ (5) $f(f(4030))=2f(f(2015))$ proof: trivially from (2),(3),(4) (6) From (1) and (5), we get $f(2015)=0$, Contradiction!!!

anantmudgal09 wrote: @khazix Well i probably am not sure about what exactly your query is but hope this might resolve. Subtract the two equations from Lemma 2 and 3 to get for all $m,n,k$ in $N$, $f(mf(n))-f(n)=f(mf(k))-f(k)$. From here we can see that the expression is constant. Still i will recheck my proof and convey it to you. The Lemma 2 equation is $f(f(n)) - f(f(k)) = n - k.$ The Lemma 3 equation is $f(mf(n)) - f(mf(k)) = n - k.$ If you subtract 2 from 3 you get $f(mf(n)) - f(f(n)) = f(mf(k)) - f(f(k))$, not $f(mf(n)) - f(n) = f(mf(k)) - f(k)$.

Oh yes. I did a typing error and you are right there is a typo. I will try to solve it again. Thanks for pointing out the error.

Tsarik wrote: Find all functions $f : \mathbb{Z}_{+} \rightarrow \mathbb{Z}_{+}$ that satisfy $f(mf(n)) = n+f(2015m)$ for all $m,n \in \mathbb{Z}_{+}$. Let $P(x,y)$ be the asserytion $f(xf(y))=y+f(2015x)$ Let $a=f(2015)>0$ and $b=f(4030)$ $P(1,x)$ $\implies$ $f(f(x))=x+a$ (and so $f(x)$ is injective). $P(1,b)$ $\implies$ $f(a+4030)=a+b$ $P(2,a)$ $\implies$ $f(2a+4030)=a+b$ And so, since injective, $a=0$, which is impossible. So no such function.

My solution: Rewrite the equation : $f(mf(n)) = n+f(2015m)$ (*) Let $m=1$ in (*) $f(f(n))=n+f(2015)$ (1) From (1) we get the injectivity of f Subtituting $n$ by $f(2015n)$ in (*) we have $f(m f(f(2015n)))=f(2015n)+f(2015m)$ (2) By changing the roles of m and n in (2) we get $f(mf(f(2015n)))=f(nf(f(2015m)))$ Thus $mf(f(2015n))=nf(f(2015m))$ (from the injectivity of f) (3) Combining (1)(3) and notice we have m,n are distinct we get $m(2015n+f(2015))=n(2015m+f(2015)) \rightarrow (m-n)f(2015)=0 $ which is a contradiction

Tsarik wrote: Find all functions $f : \mathbb{Z}_{+} \rightarrow \mathbb{Z}_{+}$ that satisfy $f(mf(n)) = n+f(2015m)$ for all $m,n \in \mathbb{Z}_{+}$. Let $P(m,n)$ be the assertion of given functional equation. Easy to see $f$ is injective. $P(1,n) \implies f(f(n))=n+c$ where $c=f(2015)$. $P(1,f(n)) \implies f(n+c)=f(n)+c$ $\implies f(n+mc)=f(n)+mc$ $P(m,mc) \implies f(mf(mc))=mc+f(2015m)=f(mc+2015m)$ $\implies f(mc)=2015+c$ $P(m,2015) \implies f(mc)=2015+f(2015m)$ $\implies f(2015m)=c$ $P(2015,2015) \implies 2015=0$ which is impossible. Hence there is no such function.

Let $P(m,n)$ denote the given equation. Then, $P(1,n)\Rightarrow f(f(n))=n+f(2015)............(1)$. $(1)$ implies that $f$ is injective. Now, $f(n+f(2015m))=f(f(mf(n)))=mf(n)+f(2015)$. Substituting $m=1$, we get $f(n+f(2015))=f(n)+f(2015)............(2)$. Now, $P(2,2015)\Rightarrow f(2f(2015))=2015+f(4030)$. Again, put $n=f(2015)$ in $(2)$. Then, $f(2f(2015))=f(f(2015))+f(2015)=2015+2f(2015)$ by $(1)$. So, $f(4030)=2f(2015)$. Next, note that by $(1)$, we have $f(f(4030))=4030+f(2015)$, while we had $f(2f(2015))=2015+2f(2015)$. Since $f(f(4030))=f(2f(2015))$, we get $f(2015)=2015$, and consequently $f(4030)=4030$. Then, we can rewrite $(1)$ as $f(f(n))=n+2015............(3)$. We can also rewrite $(2)$ as $f(n+2015)=f(n)+2015............(4)$. Now, note that due to $P(m,n),(3),(4)$, we have $f(2f(n))=n+4030=(n+2015)+2015=f(f(n))+2015=f(f(n)+2015)$. By injectivity, $2f(n)=f(n)+2015\Rightarrow f(n)=2015\forall n\in \mathbb Z_{+}$, which is impossible. Thus, there is no solution.

I'm not sure my solution is right $P(f(n);n)=>f(f^2(n))=n+f(2015f(n))$ But $f(f^2(n))=f(f(n).f(n))=f(n)+f(2015f(n))$ Therefore $f(n)+2015=n+2015=>f(n)=n$ which is impossible "Hoa nhài"

Let $c=f(2015)$, and plug in $m=1$. Then $f(f(n))=n+c$. Taking $n=2015$ implies $f(c)=c+2015$. Taking $f'$s of the previous relation, $f(n+c)=f(f(f(n)))=f(n)+c$. Induction gives $f(kc)=2015+kc$ for all integers $k\ge 1$. Now, consider $$f(cf(c))=c+f(2015c)=2016c+2015.$$We also have $$f(cf(c))=f(c^2+2015c)=2015+2015c+c^2,$$which forces $c^2=c$, or that $c=1$. But this implies that $f(n)=f(nc)=2015+nc=2015+n$. However, this means $f(2015)=4030\neq 1$, so no solutions exist.

let $P(n,m)$ the assertion $f(mf(n)) = n + f(2015m)$ $P(n,1)$$\implies f(f(n))=n+f(2015)$ (so $f$ is injective) $.......(1)$ we want to make $RHS$ symmetric to switch $(x,y)$ so we put $P(f(2015n),m)$ $\implies$ $f(mf(f2015n))=f(2015n)+f(2015m)$ $.......(2)$ from $(1)$ and $(2)$ $f(mn+mf(2015))= f(2015n)+f(2015m)$ Now switch $x,y$ $mn+mf(2015)=nm+nf(2015)$ (from the injectivity of $f$) so $f(2015)=0$ which is a contradiction So there isn't a such function

another sol

P(1,n) -> F(F(n))=n+F(2015) P(F(n),m) -> F(mF(F(n)))=F(m(n+F(2015))=F(n)+F(2015m) ** now put n=2015x (for x > m) from ** F(m(2015x+F(2015))=F(2015x)+F(2015m) => F(m(2015x+F(2015)))=F(2015x(m+F(2015))) => mF(2015)=xF(2015) => x=m

Tsarik wrote: Find all functions $f : \mathbb{Z}_{+} \rightarrow \mathbb{Z}_{+}$ that satisfy $f(mf(n)) = n+f(2015m)$ for all $m,n \in \mathbb{Z}_{+}$. First Let's Define $f(2015m)=a_m$ Given that $f$ satisfy $f(mf(n))=n+a_m---1$ Now put $m=1$ we can observe $f(f(n))=n+a_1$ clearly $f$ is Injective. And also it gives us $f(f(a_m))=a_m+a_1$ for any $m\in N$ and we can also get from equation $1$ that $f(mf(a_1))=a_1+a_m$ Hence as $f$ is Injective So $f(a_m)=mf(a_1)$ for any $m\in N$ So We get $f(a_m)=2015*m+a_1=m(2015+a_1)\implies m=0$ or $a_1=0$ Both are not possible. Hence No Such Function exist. We are Done. $\blacksquare$