41st Balkan Mathematical Olympiad

Assume that $f:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ and the polynomial $g$ with non-negative coefficients and $g(0)=0$ satisfy the conditions of the problem. For positive reals with $x>y$, we shall write $P(x,y)$ for the relation: $$f(f(x)+g(y))=f(x-y)+2y.$$ 1. Step 1. $f(x)\ge x$. Assume that this is not true. Since $g(0)=0$, $g(x)+x$ is injective on positive reals. If $f(x)<x$ for some positive real $x$, then setting $y$ such that $y+g(y)=x-f(x)$ (where obviously $y<x$), we shall get $f(x)+g(y)=x-y$ and by $P(x,y)$, $f(f(x)+g(y))=f(x-y)+2y$, we get $2y=0$, a contradiction.

2. Step 2. $g(x)=cx$ for some non-negative real $c$. We will show $\mathrm{deg}g\le 1$ and together with $g(0)=0$ the result will follow. Assume the contrary. Hence there exists a positive $l$ such that $g(x)\ge 2x$ for all $x\ge l$. By Step 1 we get $$\forall x>y\ge l:f(x-y)+2y=f(f(x)+g(y))\ge f(x)+g(y)\ge f(x)+2y$$ and therefore $f(x-y)\ge f(x)$. We get $f(y)\ge f(2y)\ge \cdots \ge f(ny)\ge ny$ for all positive integers $n$, which is a contradiction.

3. Step 3. If $c\ne 0$, then $f(f(x)+y+c^2+2)=f(x+1)+y+2c$. Indeed by $P(f(x+\frac{y}{2}+1)+\frac{cy}{2}+c,c)$, we get $$f(f(f(x+\frac{y}{2}+1)+\frac{cy}{2}+c)+c^2)=f(f(x+\frac{y}{2}+1)+\frac{cy}{2})+2c=f(x+1)+y+2c.$$ On the other hand by $P(x+\frac{y}{2}+1,\frac{y}{2}+1)$, we have: $$f(x)+y+2=f\left(f\left(x+\frac{y}{2}+1\right)+g\left(\frac{y}{2}+1\right)\right)=f\left(f\left(x+\frac{y}{2}+1\right)+\frac{cy}{2}+c\right).$$ Substituting in the LHS of $P(f(x+\frac{y}{2}+1)+\frac{cy}{2}+c,c)$, we get $f(f(x)+y+2+c^2)=f(x+1)+y+2c$.

4. Step 4. There is $x_0$, such that $f(x)$ is linear on $(x_0,\infty )$. If $c\ne 0$, then by Step 3, fixing $x=1$, we get $f(y+f(1)+2+c^2)=y+f(2)+2c$ which implies that $f$ is linear for $y>f(1)+2+c^2$. As for the case $c=0$, consider $y,z\in (0,\infty )$. Pick $x>\max (y,z)$, then by $P(x,x-y)$ and $P(x,x-z)$ we get: $$f(y)+2(x-y)=f(f(x))=f(z)+2(x-z)$$ which proves that $f(y)-2y=f(z)-2z$ and therefore $f$ is linear on $(0,\infty )$.

5. Step 5. $g(y)=y$ and $f(x)=x$ on $(x_0,\infty )$. By Step 4, let $f(x)=ax+b$ on $(x_0,\infty )$. Since $f$ takes only positive values, $a\ge 0$. If $a=0$, then by $P(x+y,y)$ for $y>x_0$ we get: $$2y+f(x)=f(f(x+y)+g(y))=f(b+cy).$$ Since the LHS is not constant, we conclude $c\ne 0$, but then for $y>x_0/c$, we get that the RHS equals $b$ which is a contradiction. Hence $a>0$. Now for $x>x_0$ and $x>(x_0-b)/a$ large enough by $P(x+y,y)$ we get: $$ax+b+2y=f(x)+2y=f(f(x+y)+g(y))=f(ax+ay+b+cy)=a(ax+ay+b+cy)+b.$$ Comparing the coefficients before $x$, we see $a^2=a$ and since $a\ne 0$, $a=1$. Now $2b=b$ and thus $b=0$. Finally, equalising the coefficients before $y$, we conclude $2=1+c$ and therefore $c=1$. Now we know that $f(x)=x$ on $(x_0,\infty )$ and $g(y)=y$. Let $y>x_0$. Then by $P(x+y,x)$ we conclude: $$f(x)+2y=f(f(x+y)+g(y))=f(x+y+y)=x+2y.$$ Therefore $f(x)=x$ for every $x$. Conversely, it is straightforward that $f(x)=x$ and $g(y)=y$ do indeed satisfy the conditions of the problem. □

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Alternative solution.

Assume that the function $f:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ and the polynomial with non-negative coefficients $g(y)=y{g}_1(y)$ satisfy the given equation. Fix $x=x_0>0$ and note that: $$f(f(x_0+y)+g(y))=f(x_0+y-y)+2y=f(x_0)+2y.$$ Assume that $g=0$. Then $f(f(x+y))=f(x)+2y$ for $x,y>0$. Let $x>0$ and $z>0$. Pick $y>0$. Then: $$2y+f(x+z)=f(f(x+y+z))=f(f(x+z+y))=f(x)+2(z+y).$$ Therefore $f(x+z)=f(x)+2z$ for any $x>0$ and $z>0$. Setting $c=f(1)$, we see that $f(z+1)=c+2z$ for all positive $z$. Therefore if $x,y>1$ we have that $f(x+y)=c+2(x+y-1)>1$. This shows that: $$f(f(x+y))=c+2(f(x+y)-1)=3c+4(x+y)-4.$$ On the other hand $f(x)+2y=c+2x+2y$. Therefore the equality $f(f(x+y))=f(x)+2y$ is not universally satisfied.

From now on, we assume that $g\ne 0$. Therefore $g$ is strictly increasing with $g(0)=0$, ${\lim }_{y\to \infty }g(y)=\infty$, i.e. $g$ is bijective on $[0,\infty )$ and $g(0)=0$. Let $x>0,y>0$ and set $u=f(x+y),v=g(y)$. From above, we have $u>0$ and $v>0$. Therefore: $$f(f(u+v)+g(v))=f(u)+2v=f(f(x+y))+2g(y).$$ On the other hand $f(u+v)=f(f(x+y)+g(y))=f(x)+2y$. Therefore we obtain that: $$f(f(x)+2y+g(g(y)))=f(f(x+y))+2g(y).$$ Since $g$ is bijective from $(0,\infty )$ to $(0,\infty )$ for any $z>0$ there is $t$ such that $g(t)=z$. Applying this observation to $z=g(g(y))+2y$ and setting $x^{\prime }=x+t$, we obtain that: $$f(f(x+t+y))+2g(y)=f(f(x^{\prime }+y))+2g(y)=f(f(x^{\prime })+g(g(y))+2y)=f(f(x+t)+g(t))=f(x)+2t.$$ Thus if we denote $h(y)=g(g(y))+2y$, then $t=g^{-1}(h(y))$ and the above equality can be rewritten as: $$f(f(x+g^{-1}(h(y))+y))=f(x)+2g^{-1}(h(y))-2g(y)=f(x)+2g^{-1}(h(y))+2y-2y-2g(y).$$ Let $s(y)=g^{-1}(h(y))+y$ and note that since $h$ is continuous and monotone increasing, $g$ is continuous and monotone increasing, then so are $g^{-1}$ and consequently $g^{-1}\circ h$ and $s$. It is also clear, that ${\lim }_{y\to 0}s(y)=0$ and ${\lim }_{y\to \infty }s(y)=\infty$. Therefore $s$ is continuously bijective from $[0,\infty )$ to $[0,\infty )$ with $s(0)=0$. Thus we have: $$f(f(x+s(y)))=f(x)+2s(y)-2y-2g(y)$$ and using that $s$ is invertible, we obtain: $$f(f(x+y))=f(x)+2y-2s^{-1}(y)-2g(s^{-1}(y)).$$ Setting $y=x_0$, we get: $$f(x)+2x_0-2s^{-1}(x_0)-2g(s^{-1}(x_0))=f(x_0)+2x-2s^{-1}(x)-2g(s^{-1}(x)).$$ Since this equality is valid for any $x>x_0$ we actually have that: $$f(x)-2x+2s^{-1}(x)+2g(s^{-1}(x))=c\text{ for some fixed constant }c\in \mathbb{R}\text{ and all }x\in {\mathbb{R}}^{+}.$$ Let $\varphi (x)=-x+2s^{-1}(x)+2g(s^{-1}(x))$. Then: $$f(f(x+y)+g(y))=f(x+y+\varphi (x+y)+c+g(y))=x+y+g(y)+\varphi (x+y)+2c+\varphi (x+y+\varphi (x+y)+c+g(y)).$$ On the other hand: $$f(f(x+y)+g(y))=f(x)+2y=x+\varphi (x)+2y+c.$$ Therefore: $$g(y)+\varphi (x+y)+c+\varphi (x+y+\varphi (x+y)+c+g(y))=\varphi (x)+y+c.$$ Noting that $\varphi$ is continuous on $[0,\infty )$, since it is sum of continuous functions, and letting $y$ tend to 0, we obtain that: $$\varphi (x)+c+\varphi (x+\varphi (x)+c)=\varphi (x).$$ Therefore $\varphi (x+\varphi (x)+c)+c=0$ and substituting in the definition of $\varphi (x)=-x+2s^{-1}(x)+2g(s^{-1}(x))$ we obtain: $$-x-\varphi (x)-c+c+2s^{-1}(x+\varphi (x)+c)+2g(s^{-1}(x+\varphi (x)+c))=0.$$ Consequently: $$c+2s^{-1}(x+\varphi (x)+c)+2g(s^{-1}(x+\varphi (x)+c))=x+\varphi (x)+c.$$ Thus: $$\varphi (c+2s^{-1}(x+\varphi (x)+c)+2g(s^{-1}(x+\varphi (x)+c)))=\varphi (x+\varphi (x)+c)=-c.$$ Finally note that $x+\varphi (x)+c=2s^{-1}(x)+2g(s^{-1}(x))+c=:u(x)$ and since $g$ and $s^{-1}$ are monotone and bijective on $[0,\infty )$, $u(x)$ exhausts $[c,\infty )$ when $x$ ranges on $[0,\infty )$. It follows that $\varphi (x)=-c$ for $x\in [c,\infty )$. It follows that for $x>\max (c,0)$: $$f(x)=x+c-\varphi (x)=x-2c.$$ In particular, since $f(x)>0$, $c\le 0$. Now for $x>\max (c,0)$ and $y>0$ we have: $$x+2y-2c=f(x)+2y=f(f(x+y)+g(y))=f(x+y-2c+g(y))=x+y+g(y)-4c.$$ Since this is valid for any $y$, we conclude $g(y)=y$ and $c=0$. Now it follows that $f(x)=x$ for $x\in (0,\infty )$. It is also straightforward to check that $f(x)=x$ and $g(y)=y$ satisfy the equality: $$f(f(x+y)+g(y))=f(x+2y)=x+2y=f(x)+2y.$$