Raja Oktovin wrote: Determine all real numbers $ a$ such that there is a function $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying \[ x + f(y) = af(y + f(x))\] for all real numbers $ x$ and $ y$. Hery Susanto, Malang Very easy. We have $ a$ not equal $ 0$. Easy prove that $ f$ be surjectivity So there exist $ m$ such that $ f(m)=0$ Let $ x=m$ we have: $ f(y)=af(y)$ So $ a=1$

yes, this is too easy, but i spot some flaws in your argument. even though it is obvious to fix, too . 1. $ a \ne 0$ 2. find that $ f$ is surjective 3. find that $ f$ is injective 4. thus there is one and only one $ x_0$ such that $ f(x_0)=0$ 5. find that $ x_0+f(y)=af(y)$ 6. if $ a \ne 1$, $ f$ is constant, impossible 7. thus $ a=1$, example $ f(x)=x$.

whole solution if $a=0$,easy to disprove if $a$ isn't zero then f is surjective it's easy to obtain $f(af(x))=\frac{x}{a}$ so $f((a-1)f(x)\equiv 0$ hence $f(x)\equiv 0$,contradiction! except for $a=1$ so the only solution is $a=1$

Here is my solution for this problem Solution $x + f(y) = af(y + f(x))$ It's easy to see that $a = 0$ doesn't satisfies the problem With $a \ne 0$: Let $y = 0$, we have: $f(f(x)) = \dfrac{x + f(0)}{a}$ So: $f$ is surjective Then there exist $b \in \mathbb{R}$ which satisfies $f(b) = 0$ Let $x = b$, we have: $(1 - a)f(y) = b$ If $a \ne 1$ then: $f(x) = c, \forall x \in \mathbb{R}, c \in \mathbb{R}$ Retry, we see that: $f(x) = c$ doesn't satisfies the problem If $a = 1$ then it's easy to see that $f(x) = x, \forall x \in \mathbb{R}$ satisfies the problem In conclusion, we have: $a = 1$ satisfies the problem

$\spadesuit \color{green}{\textit{\textbf{ANS:}}}$ $a=1$. $\clubsuit \color{red}{\textit{\textbf{Proof:}}}$ Let $P(x,y)$ denote the given assertion, we have \[P(x,f(0)): x+f(f(0))=af(f(x))\]\[P(0,f(x)): f(f(x))=af(f(x))\]so $f(f(x))=x+f(f(0))$ and \[P(x,0): x+f(0)=af(f(x))=ax+af(f(0))\]which it's now easy to see that $a$ must be $1$ in order for the equation to hold for all real $x$. In this case, $f(x)=x$ fits the given FE. $\quad \blacksquare$

Let $P(x,y)$ denote the assertion $x+f(y)=af(y+f(x))$. $P(x,f(y))\Rightarrow x+f(f(y))=af(f(x)+f(y))$ and switching $x$ and $y$, $x+f(f(y))=y+f(f(x))$. Thus, $f(f(x))=x+f(f(0))$ and $f$ must be bijective. $P(x,0)\Rightarrow x+f(0)=af(f(x))=ax+af(f(0))$ and letting $x$ vary, we see that $a$ must be $\boxed1$. In this case, $f(x)=-x$ works. Here is the solution of the resulting FE: Let $P(x,y)$ denote the assertion $x+f(y)=f(y+f(x))$. $P(x,0)\Rightarrow f(f(x))=x+f(0)$, but, as before, $f(f(x))=x+f(f(0))$, so we get $f(f(0))=f(0)$, and using injectivity, $f(0)=0$. Then $f$ is an involution. $P(f(x),y)\Rightarrow f(x+y)=f(x)+f(y)$ and any additive involution works. Let $A$ and $B$ be any two supplementary subvectorspaces of the $\mathbb Q$-vectorspace $\mathbb R$. Let $a$ from $\mathbb R\to A$ and $b$ from $\mathbb R\to B$ be the two projections of any real $x$ in $(A,B)$. We get $\boxed{f(x)=a(x)-b(x)}$.

Denote the assertion by $P(x,y).$ Clearly $a\neq 0$ then $P(x,0)$ gives $x+f(0)=af(f(x)).$ Hence $f$ is bijective. If $f(x_0)=0$ then $P(x_0,x)$ implies $f(x)(a-1)=x_0$ and since $f$ is not constant, $a=1.$

Raja Oktovin wrote: Determine all real numbers $ a$ such that there is a function $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying \[ x+f(y)=af(y+f(x))\]for all real numbers $ x$ and $ y$. Hery Susanto, Malang $$x+f(y)=af(y+f(x))...(\alpha)$$$\color{blue}\boxed{\textbf{Answer: a=1 is the only solution}}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{If a=0:}}$ $\color{red}\rule{24cm}{0.3pt}$ In $(\alpha):$ $$\Rightarrow x+f(y)=0, \forall x,y\in\mathbb{R} (\Rightarrow \Leftarrow)$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{If a=1:}}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow f(x)=x \text{ is a solution}$$$$\Rightarrow \boxed{\text{a=1 is a solution}}$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{If a}\neq \textbf{0, 1:}}$ $\color{red}\rule{24cm}{0.3pt}$ In $(\alpha):$ $$\Rightarrow \frac{x+f(y)}{a}=f(y+f(x))$$$$\Rightarrow \text{f is suryective}...(I)$$By $(I):$ $$\Rightarrow \exists b/f(b)=0$$In $(\alpha) x=b:$ $$\Rightarrow b=(a-1)f(y)$$$$\Rightarrow f(y)=\frac{b}{a-1}$$$$\Rightarrow \text{f is constant}$$In $(\alpha):$ $$\Rightarrow \text{x is constant} (\Rightarrow \Leftarrow)$$$\color{red}\rule{24cm}{0.3pt}$ $\color{blue}\rule{24cm}{0.3pt}$ $\color{blue}\boxed{\textbf{Conclusion:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$\Rightarrow \boxed{\textbf{a=1 is the only solution}}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

Raja Oktovin wrote: Determine all real numbers $ a$ such that there is a function $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying \[ x+f(y)=af(y+f(x))\]for all real numbers $ x$ and $ y$. Hery Susanto, Malang Clearly no sol for $a=0$, now i claim that $a=1$, assume FTSOC that $a \ne 1$, now let $P(x,y)$ the assertion of the following F.E. Notice that the function is surjective as $a \ne 0$, by $P(c,c)$ we get $c=0$ so $f(0)=0$, now by $P(x,0)$ we get $af(f(x))=x$ so $f$ is bijective, now note that by $P(f(x),y)$ we get that. $$f(x)+f(y)=af(y+x \cdot a^{-1}) \implies f(y+x \cdot a^{-1})=f(x+y \cdot a^{-1})$$setting $x=0$ there and then induction we can get $f(x)=f(x \cdot a^n)$ for all $n \in \mathbb Z$, now $af(f(ax))=ax$ so $f(f(x))=x$ hence $a=1$ so $f$ is involutive and additive (any function of this kind works), thus we are done .

let f constant Obviously it is wrong since $x+k = ak$ $x= k(a-1)$ implies x also constant now let $P(x,y)$ be the assertion $P(x,y) = x+f(y) = af(y+f(x))$ $P(x-f(y),f(y)) = x=af(y+f(x-f(y)))$ if $a = 0$ x constant contradiction so a is nonzero and f surjective so set some b so that $f(b) = 0$ due to the surjectivity $P(b,y) = b+f(y) = a*f(y)$ $P(b,y) = b=f(y)(a-1)$ f is non constant but b constant implies that $(a-1) = 0$ forcing $b=0$ so $a=1 , f(0)=0$