We set $x=y=0$ to get that $f(0)=f(0)+f(f(0))$, so $f(f(0))=0$. As a result there exists a real $u$, such as $f(u)=0$. Setting to the original $y=u$, we obtain that $f(2u)=f(xu)+f(0)$, so $f(xu)=f(2u)-f(0)$. If $u\ne 0$ then f is constant, let $f(x)=c$ for every real $x$. As a result $c=c+xc+c$, for every $x$, that is to say $c=0$, so one solution is $f\equiv 0$. A second case is that $u=0$, that is to say $f(0)=0$ and $f$ is injective on $0$. Setting $x=0$ to the original we get that $f(2y)=f(f(y))$ for every real $y$. Let's suppose that there exists a real $w\ne 0$, such as $f(w)=w$. Then $f(2w)=f(f(w))=f(w)=w$ and $f(4w)=f(2\cdot 2w)=f(f(2w))=f(w)=w$. Setting to the original $x=2$ and $y=w$ we get that $f(2w+2w)=f(2w)+2w+w=4w$, so $f(4w)=4w$, so combining $w=4w\Leftrightarrow w=0$, absurd! Suppose $y\ne 0$ and we set $x=\dfrac{2y}{y-f(y)}$ and since $xf(y)+2y=xy$, we get that $\dfrac{2yf(y)}{y-f(y)}+f(f(y))=0$, for every $y\ne 0$. Suppose $f(y_1)=f(y_2)$, with $y_1, y_2\ne 0$. It is direct that $f(y_1)=f(y_2)\ne 0$. Then $\dfrac{2y_1f(y_1)}{y_1-f(y_1)}+f(f(y_1))=\dfrac{2y_2f(y_2)}{y_2-f(y_2)}+f(f(y_2))\Leftrightarrow \dfrac{2y_1}{y_1-f(y_1)}=\dfrac{2y_2}{y_2-f(y_2)}\Leftrightarrow y_1=y_2$. We just proved that $f$ is injective at non-zero points and so it is injective on reals. So $f(f(y))=f(2y)\Leftrightarrow f(y)=2y$ for every real $y$, that is the second solution.

Since this is exactly the same solution as mine, I wonder if there even are reasonable alternatives.

Sylvestra wrote: Since this is exactly the same solution as mine, I wonder if there even are reasonable alternatives. dionadam wrote: We set $x=y=0$ to get that $f(0)=f(0)+f(f(0))$, so $f(f(0))=0$. As a result there exists a real $u$, such as $f(u)=0$. Setting to the original $y=u$, we obtain that $f(2u)=f(xu)+f(0)$, so $f(xu)=f(2u)-f(0)$. If $u\ne 0$ then f is constant, let $f(x)=c$ for every real $x$. As a result $c=c+xc+c$, for every $x$, that is to say $c=0$, so one solution is $f\equiv 0$. A second case is that $u=0$, that is to say $f(0)=0$ and $f$ is injective on $0$. Setting $x=0$ to the original we get that $f(2y)=f(f(y))$ for every real $y$. I guess this much part will be pretty much same for everyine. Dunno about the remaining

I think I got a significantly different solution.

I don't understand how Step 3 works.

Plugging $(x, y) \gets (0, 0)$ yields $f(f(0)) = 0$, using which by plugging $(x, y) \gets (2, f(0))$ we get $f(0) = 0$. Additionally, plugging $(x, y) \gets (0, x)$ yields $f(2x) = f(f(x))$. Assuming the existence of non-zero real $x_0$ such that $f(x_0) = 0$, by plugging $(x, y) \gets (x, x_0)$ we get $0 = f(f(x_0)) = f(2x_0) = f(xx_0)$, i.e. the first solution $f(x) \equiv 0$. Otherwise, we have $f(x) = 0 \implies x = 0$ and we assume that in the rest of the solution. The goal is to prove that such $f$ is injective. Assuming the existence of two distinct non-zero reals $x_1$ and $x_2$ such that $f(x_1) = f(x_2)$, by plugging $(x, y) \gets \left(\frac{2x_2}{f(x_1)}, x_1\right)$ and $(x, y) \gets \left(\frac{2x_1}{f(x_2)}, x_2\right)$ it follows that $x_1 = x_2$ (contradiction!), i.e. such function is injective, and we obtain the second solution $f(x) \equiv 2x$. Two students had such solution, avoiding the proof that the only fixed point of $f$ is $0$. One of them (Jozef Fülöp, Slovakia) had this shortened version and got the special recognition from Mr. Jan Kratochvíl (Dean of the Faculty of Mathematics and Physics), the Chair of the Jury. The other one (David Mikulčić, Croatia) chained nice pluggings which can be gathered in the forms $(x, y) \gets \left(\frac{2x_2}{f(x_1)}, x_1\right)$ and $(x, y) \gets \left(\frac{2x_1}{f(x_2)}, x_2\right)$, in several steps.

Sylvestra wrote: I don't understand how Step 3 works. MarkBcc168 wrote: I think I got a significantly different solution. The answers are $x\mapsto 0$ and $x\mapsto 2x$. It's easy to check that they both work. Assume that $f$ is not a zero function. Let $P(x,y)$ denote the functional equation in the problem. The proof proceeds in five steps. Step 1: $f(0)=0$ $P(x,0)$ gives $f(xf(0)) = xf(0) + [f(0)+f(f(0))]$. Thus if $f(0)\ne 0$, then $f(x) = x+c$ for some constant $c$. It's easy to check that this solution fails. Step 2: $f(a)=0\implies a=0$. Suppose that $f(a)=0$ where $a\ne 0$, then $P(x,a)$ gives $f(2a) = f(ax)$ thus $f$ is constant function. Again, this solution fails. Step 3: $f(a)=f(b)\implies f(at)=f(bt)$ for all $t\in\mathbb{R}$ Just compare $P(t,a)$ and $P(t,b)$. Step 4: $f(x)\ne f(-x)$ for all nonzero $x\in\mathbb{R}$. Assume that $f(x_0)=f(-x_0)$ for some $x_0\ne 0$, then by Step 3, $f(x)=f(-x)$ for all $x\in\mathbb{R}$. Comparing $P(x,y)$ and $P(x,-y)$ gives $$f(xf(y) + 2y) = f(xf(y) - 2y) \implies f(x+2y) = f(x-2y)$$for all $y\ne 0$. This clearly implies that $f$ is constant which fails. Step 5: $f(x)=2x$ for all $x\in\mathbb{R}$. First, $P(0,x)$ implies $f(2x) = f(f(x))$. Thus $f(2xy) = f(yf(x))$ for any $x,y\in\mathbb{R}$. Fix a nonzero real $t$ and let $k = \frac{f(t)-2t}{f(t)}$. Then $P(k,t)\implies f(kt) + kf(t) = 0$. Therefore $$f(f(kt)) = f(-kf(t)) = f(-2kt)$$which means $f(2kt) = f(-2kt)$ hence $k=0\implies f(t)=2t$ as desired. Does step 3 provide that the function is injective(if not is there way to show that is injective).

I discovered later that this was a fakesolve. Edited

Let $P(x,y)$ be the assertion $f(xf(y)+2y)=f(xy)+xf(y)+f(f(y))$. $\boxed{f(x)=0}$ is the only constant solution. Now assume that $f$ is nonconstant. If $\exists k:f(k)=0$, then assume $k\ne0$. $P\left(\frac xk,k\right)\Rightarrow f(x)=f(2k)-f(0)$ so $f$ is constant, contradiction. Such a $k$ exists since $P(0,0)\Rightarrow f(f(0))=0$. Thus $f(k)=0\Leftrightarrow k=0$. Now $P(0,x)\Rightarrow f(f(x))=f(2x)$. $P\left(1-\frac{2x}{f(x)},x\right)\Rightarrow f\left(x-\frac{2x^2}{f(x)}\right)=2x-1\forall x\ne0$ In particular, when $x=\frac12$ we get $f\left(\frac12-\frac1{2f\left(\frac12\right)}\right)=0$, so $f\left(\frac12\right)=1$. $P\left(-2,\frac12\right)\Rightarrow f(1)=2$ $P\left(1,\frac12\right)\Rightarrow f(2)=4$ $P\left(2x,\frac12\right)\Rightarrow f(2x+1)=f(x)+2x+2$ $P(2x+1,1)\Rightarrow f(4x+4)=f(2x+1)+4x+6$ so $f(4x+4)=f(x)+6x+8$ $P(x,2)\Rightarrow f(4x+4)=f(f(x))+4x+f(4)$ since $f(f(x))=f(2x)$ So we get $f(f(x))+f(4)=f(x)+2x+8$ and thus $f$ is injective. Since $f(f(x))=f(2x)$, $\boxed{f(x)=2x}$, which works.

XbenX wrote: Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that for any two real numbers $x,y$ holds $$f(xf(y)+2y)=f(xy)+xf(y)+f(f(y)).$$ Proposed by Patrik Bak, Slovakia First case when $f$ is constant: $$c=c+c \cdot x+c \implies c=0 \implies f(x) \equiv 0$$Second case when $f$ is non-constant: Let $P(x,y)$ the assertion of the given FE. $P(0,0)$ $$f(f(0))=0$$$P(x,0)$ $$f(xf(0))=(x+1)f(0) \overset{x=1}{\implies} f(0)=0$$$P(0,x)$ $$f(2x)=f(f(x))$$Now injectivity kills. If exists $a,b \ne 0$ such that $f(a)=f(b)$ i will prove that $a=b$: $P \left(\frac{2a}{f(b)}, b \right)-P \left(\frac{2b}{f(a)}, a \right)$ $$f(2(a+b))-f(2(b+a))=f \left(\frac{2ab}{f(b)} \right)-f \left(\frac{2ab}{f(a)} \right)+2a-2b+f(f(a))-f(f(b))$$Simplifying gives $a=b$. Then the solutions are: $\boxed{f(x)=2x \; \forall x \in \mathbb R}$ $\boxed{f(x) \equiv 0 \; \forall x \in \mathbb R}$ Thus we are done

$f(2x+2y)=f(\frac{2xy}{f(y)})+2x+f(2y)=f(\frac{2xy}{f(x)})+2y+f(2x)$ $f(f(x))=f(2x)$ f: injective

For storage: Denote the assertion by $P(x,y).$ Note that $P(1,0)$ gives $f(0)=0.$ If $f(z)=0$ for some $x\neq 0$ then $P(x/z,z)$ gives $f(x)=f(2z)=k.$ This $k=0$ works. $P(0,x)$ implies $f(f(x))=f(2x).$ If $f(x)\neq 0$ for all $x\neq 0.$ Then combining $P(2a/f(b),b)$ and $P(2b/f(a),b)$ gives $a=b.$ Again $P(0,x)$ gives $f(x)=2x$ and it works.

Let $f(0) = c \neq 0$. Assume $P(x,y)$ be the assertion. $P(0,y) : f(2y) = c + f(f(y))$ $P(x,0) : f(xc) = c + xc + f(c)$ so $f$ is bijective so $\exists t \in R : f(t) = 0$ $P(0,t) : f(2t) = f(xt) + f(f(t)) = f(xt) + f(0) $ so $\forall x \in R : f(xt) = f(2t) - c$ so either $f(x) =constant$ or $t = 0$. Case $1 : f(x) =constant$ Let $f(x) = k$ then $k = k + xk + k \implies k = 0$ so $\forall x \in R : f(x) = 0$ Case $2 : t = 0$ we get that $f(0) = 0$. if there exists $t' \neq 0$ such that $f(t') = 0$ then $f(xt) = f(0) = 0$ so $f$ is constant which gives contradiction. Assume there exists $x'$ such that $f(x') = x'$. $f(2x') = f(f(x')) = f(x') = x'$ and $P(2,x') : f(4x') = f(2x') + 2x' + x' = 4x'$ which gives contradiction. $P(\frac{2y}{y-f(y)},y) : \frac{2yf(y)}{y-f(y)} + f(f(y)) = 0$ Assume there exists $f(a) = f(b)$ then $P(\frac{2a}{a-f(a)},a)$ and $P(\frac{2b}{b-f(b)},b)$ implies that $a = b$ so $f$ is injective so $f(f(y)) = f(2y) \implies f(y) = 2y$. Answers $ : f(x) = 0$ and $ f(x) = 2x$.

Let $P(x,y) : f(xf(y)+2y)=f(xy)+xf(y)+f(f(y)).$ It can be easily seen that the unit function with zero applies to the problem. Therefore, we ignore this answer and assume that our function is non-zero. $P(0,\frac{x}{2}) : f(x)=f(0)+f(f(\frac{x}{2}))$ ( ) $P(0,0) : f(f(0))=0$ $P(x,0) : f(xf(0))=f(0)+xf(0)$ then put $x=1$ . then $f(0)=0$ Now we only need to say $f$ is injective ! Let $x_1 , x_2 \neq 0$ . then by $P(\frac{2x}{f(y)},y)$ and $P(\frac{2y}{f(x)},x)$ we get $f$ is injective . Now by ( ) we get $f(x)=2x$

This took a long time, also my solution is similar to #2. Answer: $f \equiv 0$ and $f(x)=2x \forall x \in \mathbb{R}$ It's easy too see that these work. Solution: Let $P(x,y)$ -denote the given assertion. First note that the only constant function that work is $f \equiv 0$ so assume $f$ is not constant Claim: $f(0)=0$ Proof: $P(x,0) \implies f(xf(0))=f(0)+f(xf(0))$ if $f(0) \neq 0$ by plugging $x \rightarrow \frac{x}{f(0)} \implies f(x)=x+f(0)$ by plugging this in into our original equation we see that there are no solution hence $\boxed{f(0)=0}$ $\square$ Claim: $f - \text{injective}$ Proof: By previous claim we saw that $f(x)=x$ doesn't work hence $f(x) \neq x \equiv f(x)-x \neq 0$. $P(\frac{2x}{x-f(x)},x) \implies f(\frac{2x^2}{x-f(x)}=f(\frac{2x^2}{x-f(x)})+\frac{2xf(x)}{x-f(x)}+f(f(x)) \implies \frac{2xf(x)}{x-f(x)}+f(f(x)) \implies$ $$\boxed{\frac{2xf(x)}{x-f(x)}=-f(f(x))} ...(*)$$ There exists $a,b \in \mathbb{R}$ such that: $f(a)=f(b)=C \neq 0$ In $(*) , x \rightarrow a \implies \frac{2af(a)}{a-f(a)}=-f(f(a)) \implies \frac{2aC}{a-C}=-f(C) ...(1)$ In $(*) , x \rightarrow b \implies \frac{2bf(b)}{b-f(b)}=-f(f(b)) \implies \frac{2bC}{b-C}=-f(C) ...(2)$ Combining $(1)$ and $(2)$ we get: $\frac{2aC}{a-C} \stackrel{(1)}{=} -f(C) \stackrel{(2)}{=} \frac{2bC}{b-C} \implies \frac{2aC}{a-C}=\frac{2bC}{b-C}$ since $C \neq 0 \implies \frac{a}{a-C}= \frac{b}{b-C} \implies -aC=-bC \implies a=b \implies \boxed{f - \text{injective}} \square$ Claim: $f(x)=2x \forall x \in \mathbb{R}$ Proof: $P(0,x) \implies f(f(x))=f(2x) \stackrel{f-\text{injective}}{\implies} f(x)=2x \forall x \in \mathbb{R} \blacksquare$ Remark: So it turns out when you get $\frac{2xf(x)}{x-f(x)}=-f(f(x))$ if you multiply everything and try to prove injectivity you get: $(a-b)(f(C)-2C)=0 \implies a=b \vee f(C)=2C , C=f(a)=f(b)$ and as far as I have seen I think this is a dead end.