Wooops, false.

Sketch of my sol during the exam, Notice that $P(y) + y$ is a polynomial which starts at $0,$ so it covers all of $\mathbb{R}^+,$ if for some $x-f(x) > 0 $ then there exists $y + p(y) = x-f(x)$ with $x > y,$ plugging said $x, y$ we get a contradiction. Now let $g(x) = f(x) - x \geq 0,$ subsituting it we get $g(g(x) + x + P(y)) + g(x) = g(x-y) + y - P(y)$ Fixing $x-y$ but letting $y \rightarrow \infty$ we get that $y-P(y) \rightarrow -\infty$ but the $L.H.S \geq 0$ contradiction, so we get that either $P(y) = 0$ or that $P(y) = y$ I missed the first case during the exam but here is the second case, namely $P(y) = y$ So we now have $g(g(x) + x + y) + g(x) = g(x-y)$ If there exist $g(c) = 0$ we can easily prove that $g=0$ constant solution (just consider larger numbers, then go down by taking $x-y$ to be small) If there does not, then $g$ is decreasing. Now fix $x,y$ and take $x'-y'=x-y$ with $x', y' \rightarrow /infty$ Comparing both sides we get that $g(x) - g(x') > x' + y' - x- y$ but the left side is bounded by $g(x)$ while the right isn't. Contradiction.

Solal wrote: Consider the sequence defined by $x_1=b$ and $x_{n+1} = \frac{x}{a+x}$. It is quite easy to show that $x_n \to 0$ when $x\to \infty$ Why is that ? If you let $y_n=\frac{1}{x_n}$ then you have $y_{n+1}=1+ay_n$ so $y_n=a^{n-1}\cdot y_1+\frac{1-a^{n-1}}{1-a} \to \frac{1}{1-a}$

Obviously $f(x)=P(x)=x$ is a solution. We'll prove that it is unique. Let $Q(x,y)$ be the given functional equation. Firstly, we'll prove that $f(x)\geq x$. Assume that for some $x_0>0$ we have $f(x_0)<x_0$. Consider the function $P(y)+y$. We know that it is continuous and it vanishes at zero. Also, $\lim_{y\to+\infty}(P(y)+y)=+\infty$. Therefore, there exists $y_0>0$ such that $P(y_0)+y_0=x_0-f(x_0)>0$. Now $Q(x_0,y_0)$ leads to $2y_0=0$, which is impossible. This inequality gives us \[f(x-y)+2y\geq f(x)+P(y)\]Assume that $\deg P\geq 2$. Then there exists $N\in\mathbb{R^+}$, such that $P(y)>2y$ for $y\geq N$. In the last inequality we take $y\geq N$ and replace $x$ with $x+y$, which means that $f(x)>f(x+y)$ for all positive reals $x$. Now in $Q(x,y)$ we take $y>N$ and an arbitrary $x>y$. Then $f(x)-x+P(y)+y>N$, so \[f(x-y)+2y=f(f(x)+P(y))=f(x-y+f(x)-x+P(y)+y)<f(x-y)\]due to the previous inequality. That's a contradiction. Therefore, $\deg P\in\{0,1\}$. Assume that $P\equiv 0$. Then \[f(f(x))=f(x-y)+2y\geq 2y\]Taking $y\to x^-$ we get that $f(f(x))\geq 2x$. Therefore, \[f(x-y)\geq f(f(x))-2y\geq 2(x-y)\implies f(x)\geq 2x,~\forall x>0\]Now, $f(x-y)+2y=f(f(x))\geq 2f(x)\geq 4x\implies f(x-y)\geq 4x-2y=2(x-y)+2x$. If we fix $x-y$ and take $x\to+\infty$, we get a contradiction. The last two results mean that $P(x)=ax$ for some $a\in\mathbb{R^+}$ Then $Q(x,y)$ becomes \[f(f(x)+ay)=f(x-y)+2y\]As before, we have $f(x-y)+2y\geq f(x)+ay\iff f(x-y)\geq f(x)+(a-2)y$, which will be denoted by $R(x,y)$. From $R(y+1,y)$ we get that \[2y+f(1)\geq f(y+1)+ay\geq (a+1)y+1\]If $a>1$, we get a contradiction for every sufficiently large $y$. Therefore, $a\leq 1$. Assume that $a<1$. We apply $R(f(x)+y,(1-a)y)$ to get that \[f(x-y)+2y=f(f(x)+ay)=f(f(x)+y-(1-a)y)\geq f(f(x)+y)+(a-2)(1-a)y\]$Q(x,\frac{y}{a})$ gives us that $f(f(x)+y)=f(x-\frac{y}{a})+\frac{2y}{a}$ Combining the last two, we get that \[f(x-y)+2y\geq f\left(x-\frac{y}{a}\right)+\left(\frac{2}{a}-a^2+3a-2\right)y\]It is easy to check that $2<\frac{2}{a}-a^2+3a-2\iff (a-1)(a^2-2a+2)<0$ which is obviously true for $a<1$. Then $f(x-y)>f(x-\frac{y}{a})$. Now, if $p>q$ are positive reals, then $x=p+\frac{(p-q)a}{1-a},~y=\frac{(p-q)a}{1-a}$ satisfy $x-y=p,~x-\frac{y}{a}=q$. Therefore, $f$ strictly increases, which implies injectivity. We can improve $R(x,y)$ to \[(2-a)y\geq f(x)-f(x-y)>0\]Taking $y\to0^+$, we get that $f$ is continuous. Now we take $y\to0^+$ in $Q(x,y)$ which gives us $f(f(x))=f(x)\implies f(x)=x$ due to injectivity. Direct check shows that this is not a solution. We are left with $a=1$, and write $I(x,y)=Q(x+y,y)$, which is\[f(f(x+y)+y)=f(x)+2y\]Now $I(f(x+y),y)$ gives us \[f(f(x+y))+2y=f(f(f(x+y)+y)+y)=f(f(x)+3y)\]From $Q(x,3y)$ we get that \[f(x-3y)+6y=f(f(x)+3y)\]Combining the last two, we obtain \[f(f(x+y))=f(x-3y)+4y\]If we replace $x$ with $x-y$ and $y$ with $2y$ in the last one, we'll get that \[f(f(x+y))=f(x-7y)+8y\]From the last two we derive \[f(x-3y)=f(x-7y)+4y\]The last one means that $f(x+y)=f(x)+y$ for every two positive integers $x,y$. If we take $x=1$, we'll get $f(y+1)=y+f(1)\implies f(y)=y+c$ for $y>1$. Now in $I(x,y)$ we fix $x$ and take $y>1$. Then \[f(x)=x+2c,~\forall x\in\mathbb{R^+}\]Taking $x>1$ means $c=0\implies f(x)=x$. This finishes the solution.

motannoir wrote: Solal wrote: Consider the sequence defined by $x_1=b$ and $x_{n+1} = \frac{x}{a+x}$. It is quite easy to show that $x_n \to 0$ when $x\to \infty$ Why is that ? If you let $y_n=\frac{1}{x_n}$ then you have $y_{n+1}=1+ay_n$ so $y_n=a^{n-1}\cdot y_1+\frac{1-a^{n-1}}{1-a} \to \frac{1}{1-a}$ Wooops sorry, it wasn't this easy finally... Idk if this is fixable, I'll try later.

The answer is $P(x) = f(x) = x$, which works. Now we prove it's the only one. Let $P(x,y)$ denote the given assertion. Claim: $f(x) \ge x$ for all $x > 0$. Proof: Suppose otherwise and some $x$ satisfied $f(x) < x$. The polynomial $P(y) + y$ can take any real value greater than $0$ and is only at most $x$ when $y < x$. So consider $P(x, y)$ with some $y < x$ satisfying $P(y) + y = x - f(x)$. We have $2y = 0 \implies y = 0$, so $f(x) = x$, absurd. $\square$ Then we have\[f(x-y) + 2y \ge f(x) + P(y) \forall x > y \ \ \ \ \ \ \ (1)\]Let $c$ be any real number. Notice that $f(x) > f(c)$ for all $x > f(c)$. Hence, for any $x > f(c)$, setting $y = x - c $ in $(1)$ gives that $f(c) + 2(x-c) \ge f(x) + P(x-c)$, so $P(x-c) < 2(x-c)$. Hence all sufficiently large numbers $k$ satisfy $P(k) < 2k$, so $P$ must be linear, meaning $P(x) = ax$ for some real $a \le 2$. Claim: $P$ cannot be the zero polynomial. Proof: Suppose otherwise. Then we have\[ f(f(x)) = f(x-y) + 2y \forall x > y\]To see that $f(f(x)) \ge 2x$, suppose otherwise. Set $y = \frac{f(f(x))}{2}$. We find that $f\left(x - \frac{f(f(x))}{2} \right) = 0$, absurd. Hence $f(f(x)) \ge 2x$ for all reals $x$. $P(x+y, y): f(f(x+y)) = f(x) + 2y$ for all positive reals $x,y$. Let $Q(x,y)$ denote this assertion. Subclaim: $f(x) \ge 2^{2^n} x$ for all nonnegative integers $n$ and positive reals $x$. Proof: We prove this by induction on $n$. Since $f(f(x+y)) \ge 2(x+y)$, we have $f(x) + 2y \ge 2(x+y)$, so $f(x) \ge 2x$, proving the base case $n = 1$. Now suppose it was true for $n = k$. We have that\[f(f(x+y)) \ge 2^{2^k} \cdot 2^{2^k} (x+y) = 2^{2^{k+1}} (x+y)\]Hence $f(x) + 2y \ge 2^{2^{k+1}} (x+y) \ge 2^{2^{k+1}} x + 2y$, so $f(x) \ge 2^{2^{k+1}} x$, as desired. $\square$ This means $f(1) \ge 2^{2^n}$, which is absurd since we can take $n$ large enough for which this is false. $\square$ Therefore $P(x) = ax$ for some $0 < a < 2$ and the equation becomes\[ f(f(x) + ay) = f(x-y) + 2y \forall x > y \]We have $f(x-y) + 2y = f(f(x) + ay) > f(x) + ay$, so\[f(x) - f(x-y) \le (2-a) y\]Let $g(x) = f(x) - x\ge 0 $. We have that\[g(x) \le g(x-y) + (1-a) y\]If $a > 1$, we would have that $g(x) \le g(x-1) - (a-1)$, so $g(x + n) \le g(x) - n (a - 1)$, absurd if we take $n$ large. Hence $a \le 1$. Claim: If $f(x) \ge Cx$ for all positive reals $x$ for some positive constant $C$, then $f(x) \ge C^2 x$ for all positive reals $x$. Proof: Suppose we had $f(x) \ge Cx \forall x > 0$. Then $P(x+y, y)$ gives that $f(f(x+y) + ay) = f(x) + 2y$ for all positive reals $x,y$, so\[ f(x) + 2y \ge C(f(x+y) + ay) \ge C^2 (x+y) + C ay\implies f(x) \ge C^2 x + y(C^2 - Ca - 2)\]Setting $y$ arbitrarily close to $0$ gives that $f(x) \ge C^2 x$. $\square$ The equation implies\[ g(g(x) + x + ay) + g(x) = g(x-y) + y(1-a) \]We see that $g(g(x) + x + ay) \le g(g(x) + x + ay - (g(x) + ay)) + (1-a) (g(x) + ay)$, which means that\[ g(x-y) + (1-a) y = g(x) + g(g(x) + x + ay) \le (3-a) g(x) + ay(1-a) \forall x > y,\]so $ (3-a) g(x) + ay (1-a) > (1-a) y$, implying that\[ g(x) > \frac{(1-a)^2}{(3-a)} y \]for all $y < x$, so taking $y$ arbitrarily close to $x$ gives that $g(x) \ge \frac{(1-a)^2}{3-a} x $ for all positive reals $x$. Claim: $a = 1 $. Proof: Suppose otherwise. Letting $c = \frac{(1-a)^2}{3-a} + 1 > 1$, we have $f(x) \ge cx \forall x > 0$, so by our earlier claim, $f(x) \ge c^{2n} x \forall x > 0$, which is a contradiction because we can take $n$ such that $c^{2n} > f(1)$. $\square$ The equation becomes\[ f(f(x) + y) = f(x-y) + 2y \forall x > y \]We have $f(x-y) + y \ge f(x)$. We have $g(x-y) \ge g(x)$, so $g$ is nonincreasing. Claim: $g$ is not lower bounded. Proof: We see that $g(f(x) + y) = f(x-y) + 2y - (f(x) + y) = f(x-y) - f(x) + y = g(x-y) - g(x)$, so $g(x-y) - g(x)$ is in the image of $g$ for all $x > y$. Taking $y$ arbitrarily close to $x$ gives the desired result (since $g$ is nonincreasing). $\square$ Now the equation implies\[g(g(x) + x + y) + g(x) = g(x-y) \forall x > y\]Hence $g(x-y) \le g(x+y) + g(x) \le 2g(x) \forall x > y$, so $g(y) \le 2 g(x) \forall y < x$, so (if $g(y) \ne 0$) taking $x$ large enough so that $g(x) < \frac{g(y)}{2}$ gives a contradiction. Hence $g$ is identically zero and we are done. @below I edited it

megarnie wrote: so either $P$ is constant or $P$ is linear with leading coefficient $1$. This implies $P(x) = 0$ or $P(x) = x$. Why ? $P$ is not in $\mathbb{Z}[X]$

Springles wrote: Now let $g(x) = f(x) - x \geq 0,$ subsituting it we get $g(g(x) + x + P(y)) + g(x) = g(x-y) + y - P(y)$ Fixing $x-y$ but letting $y \rightarrow \infty$ we get that $y-P(y) \rightarrow -\infty$ but the $L.H.S \geq 0$ contradiction, so we get that either $P(y) = 0$ or that $P(y) = y$ What if $P(y)=\frac{y}{2}$? I think from there we can only say that $\mathrm{deg} P=0$ or $1$, and if it is $1$ then $P(y)=ay$ with $a\le 1$. Assume that $P$ is a linear polynomial, and $P(x)=ax$ for some $0<a\le 1$. Then, we have \[ f(f(x)+ay)=f(x-y)+2y \]and by adding $ay$ both sides we get \[ f(f(x)) +2y = f(x-(3+a)y)+\frac{2(2+a)y}{a} \]for all $x>(3+a)y>0$. So \[ f(f(x)) = f(x-(3+a)y)+\frac{4y}{a}, \]or simply, for all $t,u>0$ the following holds \[ f(f(t+u)) = f(t)+\frac{4u}{a(3+a)}. \quad \quad (*) \] At $t=1$, we establish that $f(f(x)) = cx + d$ for all $x>1$, with $c=\frac{4}{a(3+a)}$ and $d$ as constants. Putting this into the equation $(*)$ while assuming $u$ as any constant greater than $1$, we deduce \[ f(t) = ct+d \]for all $t>0$. By quick checking, it is not difficult to see that $c=1$ and $d=0$.

Let $Q(x,y)$ be the assertion in given equation. We first assume $\deg P \geqslant 1$. $\deg P =0$ is a minor case which we will deal later. Claim 1 : $f(x) \geqslant x$ for all $x$. Proof : Suppose $x>f(x)$ for some $x$. Since the polynomial $P(x)+x$ has leading coefficient positive and $P(0)+0=0$, we can pick $y>0$ such that $P(y)+y = x - f(x)$. Note that we must have $y<x$ since $P(y)+y = x-f(x) < x$. But now $Q(x,y)$ gives a contradiction. $\blacksquare$ Claim 2 : $P(y) = ay$ for some $0<a \leqslant 1$. Proof : $Q(y+1,y) : f(1) + 2y = f(f(y+1)+P(y)) \geqslant f(y+1) + P(y) \geqslant y+1+P(y)$. Hence $f(1)-1 + y \leqslant P(y)$. Taking large $y$ proves the claim. $\blacksquare$ Claim 3 : $ \frac{f(x+\varepsilon) - f(x)}{\varepsilon}$ is lower bounded by a positive constant, for all $x,\varepsilon \in \mathbb{R}^+$. Proof : We will infact first prove 2 subclaims, showing that its infact upper bounded as well Subclaim 1 : [Upper bound] $$2-a \geqslant \frac{f(x+\varepsilon) - f(x)}{\varepsilon} \dots[\heartsuit]$$Proof : $Q(x,y) : f(x-y) + 2y = f(f(x)+ay) \geqslant f(x)+ay$. Replacing $x$ by $x+\varepsilon$ and $y$ by $\varepsilon$ and rearranging gives the desired upper bound. Subclaim 2 : [Increasing] $f(x+\varepsilon)>f(x)$ for all $x,\varepsilon>0$. Proof : Suppose $f(x+\varepsilon)\leqslant f(x)$ for some $x,\varepsilon$. Pick any $y>\varepsilon$ and subtract the two equations obtained from $P(y+x,y)$ and $P(y+x,y-\varepsilon)$. We get that $f(f(y+x)+ay) - f(f(y+x)+ay-a\varepsilon) = f(x)-f(x+\varepsilon) + 2\varepsilon$. But observe that $$\underbrace{f(f(y+x)+ay) - f(f(y+x)+ay-a\varepsilon)}_{\leqslant a\varepsilon(2-a) \text{ from }[\heartsuit]} = \underbrace{f(x)-f(x+\varepsilon)}_{\geqslant 0} + 2\varepsilon \implies a\varepsilon(2-a) \geqslant 0 + 2 \varepsilon \implies 0 \geqslant a^2-2a+2 $$this is a clear contradiction and hence the subclaim is proven! With this, we now go back to proving claim 3. For any $\varepsilon>0$, consider subtracting $P(x,y)$ from $P(x+\varepsilon,y+\varepsilon)$. We infer that $$f(f(x+\varepsilon)+a\varepsilon+ay) - f(f(x)+ay) = 2\varepsilon$$By subclaim 2, $f(x+\varepsilon) + a\varepsilon > f(x)$. Now using $[\heartsuit]$ and last equation, $$2\varepsilon = f(f(x+\varepsilon)+a\varepsilon+ay) - f(f(x)+ay) \overset{[\heartsuit]}{\leqslant} \left(f(x+\varepsilon) + a\varepsilon - f(x)\right)\cdot(2-a) \implies \frac{f(x+\varepsilon)-f(x)}{\varepsilon} \geqslant \frac{a^2-2a+2}{2-a}$$And the claim 3 is proven! $\blacksquare$ Claim 4 : $f(x)=x$ for all large enough $x$ Proof : Now we define $g(x) = f(x)-x$. Recollect that by claim1, $g(x) \geqslant 0$ for all $x$. We have the assertion $R(x,y):$ $$g(f(x)+ay)+g(x) = g(x-y) + y(1-a)$$Using the lower bound from claim 3, we can get $g(x-y)-g(x) + y(1-a) \leqslant yc$ for some constant $c$. Hence from $R(x,y)$ we infer that $$g(f(x)+ay) \leqslant yc \dots [\diamondsuit]$$Before proceeding, first observe that $f$ is surjective over all large enough positive reals (suppose $\mathbb{R}_{>M}$). This is because $Q(y+1,y) : f(1) + 2y \in \text{Im}(f)$. (And yes, this was the reason of keeping the $f(x)$ term in $R(x,y)$ ) For any $N>M+1$, and for any $y<1$, pick $x$ such that $f(x)=N-ay$ (exists due to surjectivity over $\mathbb{R}_{>M}$). By $[\diamondsuit] : g(N) \leqslant yd$. Since this is true for any $y<1$, picking arbitrarily small $y$ forces $g(N)=0$. Hence $g(x) = 0$ for all $x>M+1$ and the claim is proven. $\blacksquare$ Claim 5 : $f(x)=x$ for all $x$ and $a=1$. Proof : Pick $x>M+1$ (and hence $f(x)>M+1$ as well).In $R(x,y)$ by claim 4, the left hand side evaluates to $0+0$. Hence this forces $g(x-y)=y(1-a)=0$. From $g(x-y)=0$ : $y=x-t$ for any $t \in (0,M+1]$ implies $g(x)=0$ for all $x \leqslant M+1$ and hence combining with claim 4, $g(x)=0$ for all $x$. From $y(1-a)=0$ : We infer $a=1$. $\blacksquare$ Hence, to conclude we have $\boxed{f(x)=x \text{ for all } x \text{ and } P(x) = x}$ is the only solution.

Not too difficult, but quite long. The answer is $f(x)=P(x)=x$ only, which works. First, we deal with $P$. Claim: $f(x) \ge x$. Proof: If not, then $x+P(x) > x-f(x) > 0$, so there exists a $0<y<x$ such that $x-f(x)=y+P(y)$ by Intermediate value theorem. Hence $f(x)+P(y) = x-y \implies f(f(x)+P(y))=f(x-y) \implies 2y=0$, a contradiction. Claim: $P$ is linear. Proof: We have that \[f(x-y)+2y \ge f(x)+P(y) \ge x+P(y) \implies f(x-y) \ge x-y + P(y)-y\]If $P$ is nonlinear, then $P(y)-y$ can be arbitrarily large, fixing $x-y$ and making $y$ large yields a contradiction. Hence, we may write $P(y)=ay$ for some nonnegative real $a$, and since $P(y)-y$ must be bounded above from the proof of the above claim, it follows $a \le 1$. Now, write $g(x)=f(x)-x$. The equation becomes \[g(g(x)+x+ay)+g(x)=g(x-y)+(1-a)y\]In particular, we have \[ g(x) \le g(x-y) + (1-a)y \tag{$*$} \] Claim: For every $\varepsilon >0$, there exists $x$ such that $g(x) < \varepsilon$. In other words $g(x)$ has infinimum $0$. Proof: Consider the infinimum of $g(x)$, suppose it is $L \neq 0$. For some small $\delta$, there exists $z$ such that $g(z) < L + \delta$. Then set $x \to y+z$ for sufficiently small $y$ to get that $g(g(x)+x+ay)+g(x) = g(z) + (1-a)y < L+\delta+(1-a)y < 2L$. This means that one of $g(x)$ and $g(g(x)+x+ay)$ is smaller than $L$, contradiction. Claim: $a=1$ Proof: Substituting $x \to x+g(x)$ and $y \to g(x)$ in $(*)$ gives $g(x+g(x)) \le (2-a)g(x)$. Furthermore, substituting $x \to g(x)+x+ay$ and $y \to ay$ in $(*)$ gives that \begin{align*} g(g(x)+x+ay) \le g(x+g(x)) + a(1-a)y &\implies g(x-y) + (1-a)y \le g(x)+g(x+g(x))+a(1-a)y \\ &\implies g(x) + g(x+g(x)) \ge (1-a)^2y + g(x-y) \ge (1-a)^2y\\ &\implies (3-a)g(x) \ge (1-a)^2y \end{align*}for all $y<x$. Hence $g(x) \ge \frac{(1-a)^2}{3-a} x$. This contradicts the previous claim unless $a=1$. Finish: If $a=1$ then the original equation in terms of $g$ implies $g(x) \le g(x-y)$, i.e $g$ is decreasing. This, combined with $g(x)$ having infinimum $0$, implies that for each $\varepsilon$, $g(x) < \varepsilon$ for all sufficiently large $x$. Hence, in the original equation, fixing $x-y$ at some arbitrary value and making $x$ very large, we have $g(x-y) < \epsilon$ for all $\epsilon>0$, which means $g$ is the zero function, so $f(x)=x$ and we are done.

Very Nice problem! Claim 1: $f(x)\geq x$

Claim 2: $P = cx$ where $c \in [0,1]$

Let $g(x)=f(x)-x$, then the FE becomes: $$g(g(x)+x+cy)+g(x)=g(x-y)+(1-c)y$$and $P(x,y)$ denote this assertion. Claim 3: $\inf(g) =0$

Claim 4: $c=1$

Claim 4 implies that $g$ is decreasing, and that combined with Claim 3, $g \equiv 0$ and $f(x) \equiv x$ and they clearly work. $\blacksquare$

mathuz wrote: Springles wrote: Now let $g(x) = f(x) - x \geq 0,$ subsituting it we get $g(g(x) + x + P(y)) + g(x) = g(x-y) + y - P(y)$ Fixing $x-y$ but letting $y \rightarrow \infty$ we get that $y-P(y) \rightarrow -\infty$ but the $L.H.S \geq 0$ contradiction, so we get that either $P(y) = 0$ or that $P(y) = y$ What if $P(y)=\frac{y}{2}$? I think from there we can only say that $\mathrm{deg} P=0$ or $1$, and if it is $1$ then $P(y)=ay$ with $a\le 1$. Assume that $P$ is a linear polynomial, and $P(x)=ax$ for some $0<a\le 1.$ You’re right, this is the case I missed during the exam.

I think no posted a solution which uses my idea (for the last part) so I'll post it. Let $P_1(x,y)$ be the given FE. If there exist $x>y>0$ so that $f(x)+P(y)=x-y$ then $P_1(x,y)$ yields $y=0$ which is absurd. Hence, $y+P(y)\neq x-f(x)$ for all $x>y>0.$ However, \[\lim_{y\to0^+}y+P(y)=0,\quad\lim_{y\to x^-}y+P(y)=x+P(x)\geqslant x,\]so for any $\lambda\in(0,x)$ there exists $y\in(0,x)$ such that $y+P(y)=\lambda.$ Consequently, $x-f(x)\not\in (0,x)$ which implies $f(x)\geqslant x.$ Using this on the LHS of $P_1(y+1,y)$ we get $P(y)\leqslant y+c$ for some constant $c{}$ and any $y\geqslant 0.$ Therefore, $\deg P\leqslant 1$ and taking in consideration $P(0)=0$ we get $P\equiv 0$ or $P(x)=cx$ for some constant $c>0.$ We begin with the case $P\equiv 0$ giving us the FE $f(f(x))=f(x-y)+2y$ denoted by $P_2(x,y).$ Choose arbitrary $a,b>0$ and compare $P_2(a+b,a)$ and $P_2(a+b,b)$ to get that $f(a)-2a=f(b)-2b.$ Therefore, $f(x)-2x$ is constant, so $f(x)=2x+c$ for some constant $c.$ It can be easily checked that this doesn't work. Now, let's treat the case $P(x)=cx$ with $c>0.$ We get the FE $f(f(x)+cy)=f(x-y)+2y$ denoted by $P_3(x,y).$ Recall the condition $x>y>0.$ Choose $x>y(c+1)/c$ and look at $P_3(f(x)+y(c+1),y).$ In the LHS:\begin{align*}f\left(f\left(f(x)+(c+1)y\right)+cy\right)&=f\left(f\left(f(x)+c\cdot \frac{y(c+1)}{c}\right)+cy\right) \\ &=f\left(f\left(x-\frac{y(c+1)}{c}\right)+y\left(2+c+\frac{2}{c}\right)\right).\end{align*}We used $P_3(x,y(c+1)/c)$ for the second equality. The RHS has a much simpler form; we use $P_3(x,y)$ to get:\[f(f(x)+cy)+2y=f(x-y)+4y.\]Let $x=y(c+1)/c+a$ for some arbitrary $a>0.$ Putting togheter the former and latter expressions we finally get:\[f\left(a+\frac{y}{c}\right)+4y=f\left(f(a)+y\left(2+c+\frac{2}{c}\right)\right)\geqslant f(a)+4y+\frac{y}{c}.\]We've obtained the inequality by using $f(x)\geqslant x$ and the am-gm inequality on $c+1/c\geqslant 2.$ Evidently, $y/c$ can be any real number $b>0.$ Therefore, $f(a+b)\geqslant f(a)+b$ for any $a,b>0.$ Using this on the RHS of $P_3(x,y)$ we have\[f(f(x)+cy)=f(x-y)+2y\leqslant f(x+y).\]In particular, from $f(a+b)\geqslant f(a)+b$ it follows that $f$ is increasing, so we get $f(x)\leqslant x+y(1-c)$ for any $x>y>0.$ By taking $x>y\to0^+$ we may infer that $f(x)\leqslant x$. Combining this with $f(x)\geqslant x$ we get $f(x)=x.$ Checking easily yields $P(x)=x$ as well. To conclude, $f(x)=P(x)=x$ is the only solution.

Here's my geometric solution from contest where the only analytic result are the linear bounds. Allegedly it was mentioned in special prize discussions, but I don't know if that's true.

@above ...wow. I think that the allegations are indeed true. My solution for this problem was included in such discussions, but eventually lost the jury vote after comparison to other solutions of problem 4, such as your own one. Although I haven't fully read it yet, I think that your solution is unique enough to be awarded a special prize

As I probably will not have time and nerves to dive deeply into this problem, can anyone just comment on how similar it is to Iran 3rd Round 2021. (I am not pretending that someone has stolen a problem or sth, but just remember during the competition that I have seen this; one of the contestants also knew about the existence of the Iranian problem, but did not help him at all since he did not remember distinguishing ideas from it.)

Assassino9931 wrote: [...] Can anyone just comment on how similar it is to Iran 3rd Round 2021. When I saw the BMO problem for the first time, the Iran problem immediately came to mind (because FE's with polynomials are not that common, I guess) but other than having the same "flavour" I think the problems are fairly different. In the Iran problem, I think that it is crucial to study the image and bijectivity of the function, which can also be done for the BMO problem, but it turns out to be more convoluted than necessary. They did a good job with this FE (in my opinion) because it does not rely on analysis as much as other $\mathbb{R}^+$ polynomial FE's do (yes, I know I used limits, but they are avoidable). Every single solution on the Iran problem thread follows (mostly) the same path of finding the image of the function and proving bijectivity. On the other hand, for the BMO problem, hardly anybody used these ideas.

Is the problem solvable if we drop the condition that $P(0)=0$?

Orestis_Lignos wrote: Is the problem solvable if we drop the condition that $P(0)=0$? Yes, my solution still works. In the beggining, we get $f(x)\geqslant x-P(0)$ rather than $f(x)\geqslant x$ but this still yields $\deg P\leqslant 1.$ Then, for a constant polynomial, the solution works exactly the same. For $\deg P=1$ let $P(x)=cx+P(0).$ Our FE is $f(f(x)+cy+P(0))=f(x-y)+2y$ with $0<y<x.$ Denote it by $P(x,y).$ Take $x>y(c+1)/c$ and look at $P(f(x)+(c+1)y+P(0),y).$ In the LHS we have \begin{align*}f\left(f\left(f(x)+(c+1)y+P(0)\right)+cy+P(0)\right)&=f\left(f\left(f(x)+c\cdot \frac{y(c+1)}{c}+P(0)\right)+cy+P(0)\right) \\ &=f\left(f\left(x-\frac{y(c+1)}{c}\right)+y\left(2+c+\frac{2}{c}\right)+P(0)\right).\end{align*}For the RHS we have $f(f(x)+cy+P(0))+2y=f(x-y)+4y.$ Let $x=y(c+1)/c+a$ for some arbitrary $a>0.$ Using am-gm like before and $f(x)\geqslant x-P(0)$ which cancels out the extra $P(0){}$ we have\begin{align*}f\left(a+\frac{y}{c}\right)+4y=f\left(f\left(a\right)+y\left(2+c+\frac{2}{c}\right)+P(0)\right)\geqslant f(a)+\frac{y}{c}+4y.\end{align*}Now we finish analogously to the previous solution.

I think dropping the condition just makes the linear bounds somewhat less tight in my solution. So it should work just fine either way.

falantrng wrote: Let $\mathbb{R}^+ = (0, \infty)$ be the set of all positive real numbers. Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ and polynomials $P(x)$ with non-negative real coefficients such that $P(0) = 0$ which satisfy the equality $f(f(x) + P(y)) = f(x - y) + 2y$ for all real numbers $x > y > 0$. I am going to solve this from $P(y)=ay$ with $0<a<=1$ and then: We have that: $f(f(x)+ay)=f(x-y)+2y$ for $x>y>0$ Taking $x\rightarrow f(x)+az$ we get: $f(f(x-z)+2z+ay)=f(f(x)+az-y)+2y=f(f(x)+a(z-\frac{y}{a}))+2y=f(x-z+\frac{y}{a})+2z-\frac{2y}{a}+2y$ For $x>y,z$ and $z>\frac{y}{a}$ If $x\rightarrow x+z$ we get: $f(f(x)+2z+ay)==f(x+\frac{y}{a})+2z-\frac{2y}{a}+2y$ when $z>\frac{y}{a}$ For $z\rightarrow \frac{f(z)}{2},y\rightarrow ya$ we get: $f(f(x)+f(z)+a^2y)=f(x+y)+f(z)+2ya-2y$ for $f(z)>2y$ Using the symmetry of $x,z$ we get that: $f(x+y)-f(x)=f(z+y)-f(z)\Rightarrow f(x+y)-f(x)=g(y)\Rightarrow f(x+y)=f(x)+g(y)$ for $f(x)>2y$. Because $f(x)>=x$ we can say that it is true for $x>2y$. Now fix $x,y$ and sellect a big $z$ then we have: $f(z)+g(x+y)=f(z+x+y)=f(z+x)+g(y)=f(z)+g(x)+g(y)\Rightarrow g(x+y)=g(x)+g(y)$ Now for $x,y$: $f(x+y)=f(x+n*\frac{y}{n})=f(x+(n-1)*\frac{y}{n})+g(\frac{y}{n})=...=f(x)+ng(\frac{y}{n})=f(x)+g(y)$ for every $x,y$ By the symmetry we get that $g(x)=f(x)+c$ and so $f(x+y)=f(x)+f(y)+c$ Which by Cauchy-type lemma we have that $f(x)=ax+b$ and esily we get that $f(x)=P(x)=x$ is the only solution.

Let $P(x,y)$ as the assertation $f(f(x) + P(y)) = f(x - y) + 2y$ Claim 1: $f(x)\geq x$

Claim 2: $deg$ $P\leq1$

Claim 3: $deg$ $P\neq0$

Claim 4: $c\leq 1$

Claim 5: $f$ is linear

Claim 6: $f(x)=x, P(x)=x$

Here is a solution I came up with recently. It repeats some of the steps, but differs in the crucial one. It consists of three steps. The first two are present in many of the above solutions 1) $f(x)\ge x$, 2) $P(x)=ax, a\in[0,1]$. The third step is to prove that $f(x)-x\le \varepsilon, \forall x\in\mathbb{R}^+$ holds for any $\varepsilon>0$. Step 1. We see that $f(x)\equiv x$ and $P(x)\equiv x$ are solutions. So, it is helpful to set $g(x):=f(x)-x$ and try to prove $g\equiv 0$. With respect to $g$ the equation becomes $$\displaystyle g\big(g(x) +x +P(y) \big)+ g(x)=g(x-y)+y -P(y)\qquad (*)$$We try choosing two numbers $x,y, x>y>0$ in such a way that $$g(x) +x +P(y)=x-y \qquad (1)$$Suppose that this can be done. Then $g(x-y)+g(x)=g(x-y)+y-P(y)$, so $g(x)=y-P(y)$, but this contradicts $(1)$ . It remains to see that in case $g(x)<0$ we can always find $y$ such that $x>y>0$ and $(1)$ is satisfied. Indeed, $(1)$ is equivalent to $-g(x)= y+P(y)$ and the left hand side is positive. Since $y+P(y)$ is continuous, $0+P(0)=0$ and $x+P(x)>-g(x)= x-f(x)$, we conclude that there is $y, 0<y<x$ that satisfies $(1).$ Thus, $g(x)<0$ is impossible, therefore $g(x)\ge 0\,,\,\forall x\in\mathbb{R}^+.$ Step 2. $P(x)=ax$, where $a\in [0,1]$. Setting $z=x-y$, from $(*)$ we get $$\displaystyle g\big(g(z+y) +z+y +P(y) \big) \le g(z)+y -P(y)\qquad (2)$$From here, it easily follows $P(y)=ay$ where $a\in [0,1]$. Indeed, otherwise $\lim_{y\to\infty} (y-P(y))=-\infty$ and fixing $z$ it would contradict $(2)$. Actually this step is not needed. Step 3. For any $\varepsilon>0$, it holds $f(x)-x\le \varepsilon, \forall x\in\mathbb{R}^+$. From the above steps, we get $$\displaystyle g\big(g(z+y) +z+(1+a)y \big)+ g(z+y)=g(z)+(1-a)y \qquad(3),$$which holds for all $y,z>0$. Here is the idea loosely speaking. We first prove that $g(z)$ takes values as close to $0$ as we want. Then we take some $z_1$ with $0\le g(z_1)<\varepsilon$, where $\varepsilon>0$ is small and we choose a small $y>0$. Note that the right-hand-side of $(3)$ is small. This means that the both terms in the left-hand-side of $(3)$ are small, hence at least one of them will be less than half of the RHS of $(3)$, that is, smaller than $\varepsilon$. In this way we can construct a sequence of numbers $0<z_1<z_2<\dots$, close to one another, but not too close, such that $g(z_i),i=1,2,\dots$ are all less than $\varepsilon$. Again, by $(3)$, if $g(z)$ is a small number and $y$ is small then $g(z+y)$ is also small. Therefore, for all $z\in (z_i,z_{i+1})$, $g(z)$ is less than some small number (that depends only on $\varepsilon.$) Further, it easily follows the same holds for all $z\in\mathbb{R}^+$. So, there is no option other than $g(z)\equiv 0$. One can see the details in this blog post.

Strudan_Borisov wrote: We can improve $R(x,y)$ to \[(2-a)y\geq f(x)-f(x-y)>0\]Taking $y\to0^+$, we get that $f$ is continuous. Now we take $y\to0^+$ in $Q(x,y)$ which gives us $f(f(x))=f(x)\implies f(x)=x$ due to injectivity. Direct check shows that this is not a solution. Sorry I don't understand this part. Can you explain the reason that taking $y \to 0^+$ give us that $f$ is continuous?

Denote $P(x,y)$ the assertion of the F.E.: $f(f(x+y)+P(y))=f(x)+2y, \; \forall x,y \in \mathbb R^+$ First on the original one if you had $x>f(x)$ then note that since $P(z)$ is non-decreasing and continuos we have that so is $P(z)+z$ (this one is strictily increasing in fact) so we can choose $f(x)+P(y)+y=x$, clearly $x>y$ here so it's safe and thus we would have $2y=0$, contradiction!. Therefore $f(x) \ge x$ for all $x \in \mathbb R^+$. Now note that $f(x)+2y \ge f(x+y)+P(y)$ which shows that $P$ is linear, as if had degree greater or equal than $2$ then fixing $x$ and making $y$ be arbitrarily large gives a size contradiction, therefore $P(z)=az$ (because $P(0)=0$), for a constant $2 \ge a \ge 0$. Now our new F.E. is $f(f(x+y)+ay)=f(x)+2y$ for all $x,y \in \mathbb R^+$. Again notice that $f(x)+2y \ge f(x+y)+ay \ge x+y+ay$ which means that $1 \ge a$, now we prove that $a \ne 0$. Suppose otherwise then $f(f(x+y))=f(x)+2y$ which by symetry implies that $f(x)+2y=f(y)+2x$ so $f(x)=2x+c$ for some constant $c \ge 0$, but then replacing back gives $4(x+y)+3c=2(x+y)+c$ contradiction!, thus $a \ne 0$. Now from $P(f(x+y),ay)$ we get that $f(f(x+y)) \ge f(x)+y+(a-1)^2y \ge f(x)+y$ for all $x,y \in \mathbb R^+$. $P(f(x)+ay,y)$ gives $f(f(f(x)+(a+1)y)+ay)=f(x-y)+4y$ for $x>y$, now in fact if we wanted to unlock LHS as well we let $x>\frac{(a+1)y}{a}$ and thus we have: \[ f(x-y)+4y=f \left(f \left(x-\frac{(a+1)y}{a} \right)+ay+\frac{2(a+1)y}{a} \right) \ge f \left(x-\frac{(a+1)y}{a} \right)+y \left(\frac{1}{a}+2+a+\frac{1}{a} \right) \overset{AM-GM}{\ge} f \left(x-\frac{(a+1)y}{a} \right)+4y+\frac{y}{a} \]Now here go $(x,y) \to (ax,ay)$ and also note that we can now split this into the choice of $ax-ay=u$ and $y=v$ (which covers everything $u>v$ by choosing a suitable $x,y$) thus we get that $f(u) \ge f(u-v)+v$ for all $u>v$ or equivalent to $f(s+t) \ge f(s)+t$ for all $s,t \in \mathbb R^+$. Notice that the things we got give $f(x) \ge \lim_{y \to 0} f(x-y)$ and $\lim_{y \to 0} f(x+y) \ge f(x)$, but from the original assertion we have that $f(x-y)+2y \ge f(x)+cy$ for $x>y$ so setting $y \to 0$ here gives $\lim_{y \to 0} f(x-y) \ge f(x)$ and on $P(x,y)$ by using the ineq we get that $f(x)+2y \ge f(x+y)+cy$ so by letting $y \to 0$ we get $\lim_{y \to 0} f(x+y) \le f(x)$, but now notice that combining all of these we get limit from right and left all being equal to $f(x)$, therefore $f$ is a continuos function. Now take $y \to 0$ on $P(x,y)$, then we get that $f(f(x))=f(x)$ and note that $f$ is surjective for arbitarily large values wherefore $f(x)=x$ for all $x>N$, now just set $t>N$ to get $s+t \ge f(s)+t$ for all $s \in \mathbb R^+$ which gives that $f(s) \ge s \ge f(s)$ therefore $f(x)=x$ for all $x \in \mathbb R^+$ thus we are done .