goodar2006 wrote: $g(x)$ is a polynomial of degree at least $2$ with all of it's coefficients positive. Find all functions $f:\mathbb R^+ \longrightarrow \mathbb R^+$ such that for all $x,y\in \mathbb R^+$ $f(f(x)+g(x)+2y)=f(x)+g(x)+2f(y)$ It's rather easy to show that $f(x)=x$ is the only continuous solution. Could you kindly confirm us that the problem must be solved without continuity constraint ?

pco wrote: Could you kindly confirm us that the problem must be solved without continuity constraint ? Yes , It's can be solved without any other restrictions .

Does $0\in \mathbb{R_+}$ in this problem?

PhantomR wrote: Does $0\in \mathbb{R_+}$ in this problem? I only translated what was written on the exam paper, but I think the answer to your question is no.

goodar2006 wrote: $g(x)$ is a polynomial of degree at least $2$ with all of its coefficients positive. Find all functions $f:\mathbb R^+ \longrightarrow \mathbb R^+$ such that for all $x,y\in \mathbb R^+$ $f(f(x)+g(x)+2y)=f(x)+g(x)+2f(y)$ Let $P(x,y):f(f(x)+g(x)+2y)=f(x)+g(x)+2f(y)$. We will show that for some $c>0$ and $M>0$, for all $x>M$ we have $f(x)+c=f(x+c)$.

Let $d=g(x_0 +c)-g(x_0)$, for some $x_0 >M$. We show that $f(r)+d=f(r+d)$ for all reals $r$.

Since some interval $[k, \infty )$ is contained in $ \{ g(x+c)-g(x), x>M \}$, we have $f(r)+d=f(r+d)$ for any $d \geq k$. Consider any $p>0$. Since $f(r+p)+k=f(r+p+k)=f(r)+p+k$, we obtain that $f(r+p)=f(r)+p$ for any $r,p>0$. Thus $f(x)=x+a$ for some $a$. A quick check shows that $f(x)=x$ is the only solution. My thoughts:

Congrats! Awesome! During the exam, only three people (out of sixteen!) could solve it!

put $h(x)=f(x)+g(x)$.because $g$ is a polynomial so $h$ is not constant now if $r,s$ are two positive reals such that we have $h(r)-h(s)=2T$ for some positive $T$ we have:$h(s)+2y+2T=h(r)+2y$ so with take an $f$ we have $f(y+T)=f(y)+T$ so for natural $n$ ,$f(y+nT)=f(y)+nT$ now for an arbitrary $x$ choose $y$ such that $y+h(x)=nT$ for some $n\geq2$ so we have $2f(y)+h(x)=f(2y+h(x))=f(y+nT)=f(y)+nT$ so $f(y)=y$. now because $h(x+T)-h(x)=T+g(x+T)-g(x)$ so there exist $c$ such that it is surjective after $c$ so for any $l\geqq$ we have $f(y+l)=f(y)+l$ now with use the fact that $f$ has a fixed point we have $f(y)=y$ for any positive $y$.

Let $P(x,y)$ be the given assertion. Obviously, $h(x)=f(x)+g(x)$ is not a constant function, Outline of my solution: Take some $x, y$ such that $h(x)>h(z)$ and take $d=\frac{h(x)-h(z)}{2}$ Then, $P(x,y)$ and $P(z,y+d)$ imply $f(y+d)=f(y)+d$ So, the function $f(x)-x$ is periodic and hence bounded. That is, $|f(x)-x|<M$ for some real $M$. But, $f(y)-y=\frac{f(2y+h(x))-(2y+h(x))}{2}=\frac{f(4y+3h(x))-(4y+3h(x))}{4}$ $=\cdots =\frac{f(2^n y + (2^n - 1)h(x)) - (2^n y + (2^n - 1)h(x)) }{2^n}$ So, $|f(y)-y|<\frac{M}{2^n}$ for large $n$. So, $f(y)=y$, done!

bappa1971 wrote: So, the function $f(x)-x$ is periodic and hence bounded. "periodic implies bounded" is true only if you have continuity (there are a lot of periodic unbounded non continuous functions) And we dont have.

Pedram-Safaei wrote: now because $h(x+T)-h(x)=T+g(x+T)-g(x)$ so there exist $c$ such that it is surjective after $c$ so for any $l\geqq$ we have $f(y+l)=f(y)+l$ now with use the fact that $f$ has a fixed point we have $f(y)=y$ for any positive $y$. Can anyone gives a detailed explanation for these lines please?

This is a really nice problem! goodar2006 wrote: Let $g(x)$ be a polynomial of degree at least $2$ with all of its coefficients positive. Find all functions $f:\mathbb R^+ \longrightarrow \mathbb R^+$ such that \[f(f(x)+g(x)+2y)=f(x)+g(x)+2f(y) \quad \forall x,y\in \mathbb R^+.\] Proposed by Mohammad Jafari Let $h(x) \overset{\text{def}}{:=} f(x)-x$, $p(x) \overset{\text{def}}{:=} x+g(x)$ and $t(x) \overset{\text{def}}{:=} h(x)+p(x)$ for all $x>0$; then we obtain $$h(h(x)+p(x)+2y)=2h(y)$$for all $x,y>0$. Swap $x$ with a variable $z>0$; then $$h(h(x)+p(x)+2y)=h(h(z)+p(z)+2y)=2h(y)$$for all $x,y,z>0$. Notice that if $h$ is injective, then $f(x)=-g(x)+c$ which is false since $g(x) \rightarrow \infty$ as $x \rightarrow \infty$ while $f(x)>0$ always holds. Pick $a,b$ with $t(a) \ne t(b)$ and set $c=|t(a)-t(b)|$ then for all $x>N=\min(t(a), t(b), 0)$ we have $h(x)=h(x+c)$. Now put $y=\varepsilon$ and $x$ to be large enough so that $g(x)>N$ (hence $t(x) \ge p(x)-x=g(x)>N$), and substitute $y$ with $\varepsilon+c$; then $f(c+\varepsilon)=f(\varepsilon)$. Thus, $h$ is periodic with period $c$; i.e. $h(x+c)=h(x)$ for all $x>0$. Now notice that $h(x+c)+g(x+c)-h(x)-g(x)=g(x+c)-g(x)$ is a period of $h$, for all $x>0$. Thus, all points in an entire interval are periods of $h$ proving $h$ is the constant function. Clearly, $h(h(1)+p(1)+2)=2h(1)$ shows that $h \equiv 0$. Hence $f$ is the identity function and it clearly works! $\blacksquare$

The answer is $f(x) \equiv x$, which clearly works. Let $P(x,y)$ be the given assertion, and $h(x) = f(x) + g(x)$. Let set $$S = \{s \in \mathbb R^+ : \exists ~ x_1,x_2 \in \mathbb R^+ \text{ such that } s = h(x_1) - h(x_2) \}$$ Claim 1: For any $s \in S$, we have $f(y+s) = f(y) + s ~~ \forall ~ y \in \mathbb R^+$. Proof: If $s = h(x_1) - h(x_2)$, just compare $P(x_1,y)$ and $P(x_2,y+s)$. $\square$ Claim 2: Function $h$ is non-constant, i.e. set $S$ is non-empty. Proof: Note that as $x$ becomes large, $g(x)$ also becomes arbitrarily large. As $f(x) > 0$, so $h(x) = f(x) + g(x)$ becomes arbitrarily large, which clearly proves our Claim. $\square$ Claim 3: There is constant $M \in \mathbb R^+$ such that all numbers $\ge M$ lie in $S$. Proof: Fix any $s \in S$. Using Claim 1 we have $$ S \ni h(x+s) - h(x) = f(x+s) - f(x) + (g(x+s) - g(x)) = s + t(x) $$where $t(x) := g(x+s) - g(x)$. Note $t$ is a non-constant polynomial and all its coefficient are positive. We easily have that all large numbers are in range of $t$, which proves our Claim. $\square$ Combining Claim 1 and Claim 3 we get $$ f(x+r) = f(x) + r ~~~ \forall ~ x \in \mathbb R^+ , r \ge M \qquad \qquad (\star)$$ Claim 4: We have $f(x) \equiv x+c$ for some constant $c$. Proof: Fix any $0 < a < b$. It suffices to show $$ f(b) - f(a) = b-a $$Pick any $r \ge M$. Using $(\star)$ we get $$ f(b) - f(a) = \left( f(b+r) - f(a) \right) - \left(f(b+r) - f(b) \right) = (b+r-a) - r = b-a $$which proves our Claim. $\square$ Lastly, by look at $P(x,y)$, it isn't hard to see that we must have $c=0$, completing the proof. $\blacksquare$