I solved it like this: $x=1$ gives $f(f(y)+1)=y+f(1)$. Now, $x=f(x)+1$ gives $f(f(x)f(y)+f(x)+f(y)+1)=f(x)y+y+x+f(1)$. Since the LHS of this new equation is symmetric about $x$ and $y$, we get $f(x)y=f(y)x$ for all $x,y$. So the solution is $f(x)=cx$, by substituting in, we get $c^2xy+cx=xy+cx$, from which $c=\pm 1$

fattypiggy123 wrote: Find all functions from the reals to the reals satisfying \[f(xf(y) + x) = xy + f(x)\] $f$ is injective ( This is obvious). now from $P(x,\frac{-f(x)}{x}) $ we have : $f(\frac{-f(x)}{x})=-1\: \: \forall x\neq 0$ so ($f$ is injective) $\frac{-f(x)}{x}=c\rightarrow f(x)=kx$

Letting $x = 1$ we see that $f(f(y) + 1) = y + f(1)$, which means that $f$ is surjective. Now, let $a$ such that $f(a) = 0$. Then, letting $y = a$ in the original equation we get $f(x) = ax + f(x) \Rightarrow ax = 0$, so, we must have $a = 0$, hence $f(0) = 0$. Now, let b such that $f(b) = -1$. Letting $y = b$ in the original equation, we get $f(0) = bx + f(x) \Rightarrow f(x) = -bx$. Substituting this in the original equation, we get $-b(x(-by) + x) = xy -bx \Rightarrow (b^2 -1)xy = 0$, so, we must have $b^2 = 1$., so $b = \pm 1$. It can be verified that both $f(x) = x$ and $f(x) = -x$ satisfy the original equation.

Answer: $f(x) = x$ or $f(x) = -x$. Let $P(x,y)$ denote the given assertion. $P(\text{constant} , y) \implies f$ is surjective. For injectivity assume $f(a) = f(b)$ $\implies xf(a) = xf(b) \implies xf(a) + x = xf(b) + x \implies f(xf(a) + x) = f(xf(b) + x)$ $\implies ax + f(x) = bx + f(x) \implies a=b$ (where $x \neq 0$) $\implies f$ is injective except at $0$. $P(x,0) \to f(xf(0) + x) = f(x) \implies \implies xf(0) + x = x \implies f(0) = 0$. Let there be $a$ such that $f(a)=-1$. $P(x,a) \to f(0) = ax + f(x) \implies f(x) = -ax$. Plugging this back in we get $a^2 = 1 \implies a = \pm 1$ We will prove that $f(x) = x \quad \textbf{or} \quad f(x) = -x$ works. Now assume for contrary that there exists $a,b \neq 0$ such that $f(a) = a$ and $f(b)=-b$. $P(a,b) \to f(-ab + a) = ab + f(a)$ Case 1: $(-ab + a) = ab + f(a) \implies -ab = ab$ which is absurd. Case 2: $-(-ab + a) = ab + f(a) \implies -a = f(a) = a$ which is absurd as well, and we are done.

If $\exists k:f(k)=0$, then $P(1,k)\Rightarrow k=0$. Let's force the RHS to be zero. $P(x,-f(x)/x)\Rightarrow f(xf(-f(x)/x)+x)=0\Rightarrow f(-f(x)/x)=-1$ Now $P(1,x)\Rightarrow f(f(x)+1)=x+f(1)$, so $f$ is bijective. Thus $-f(x)/x=-c$ for some constant $c$, hence $f(x)=cx$. Testing, we get $c\in\{-1,1\}$, hence the two solutions $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$.

I discussed this problem on my channel: Link: Video Solution

fattypiggy123 wrote: Find all functions from the reals to the reals satisfying \[f(xf(y) + x) = xy + f(x)\] Let $P(x,y)$ the assertion of the given FE. $P(f(x),-1)$ $$f(f(x)f(-1)+f(x))=0 \implies f(x)(f(-1)+1)=c \; \text{where} \; c \; \text{is a constant such that} \; f(c)=0$$Note that if $f(x)$ is a solution hence $-f(x)$ its another solution so assume that if $x>0$ then $f(x)>0$ , or the inverse , if $x<0$ then $f(x)<0$. It is easy to see that there is no constant solucions and hence $c=0$ and $f(-1)=-1$. $P(x,-1)$ $$f(0)=f(x)-x \implies \; \text{since} \; f(0)=0 \; \text{then} \; f(x)=x \: \forall x \in \mathbb R$$And then the solutions are: $\boxed{f(x)=\pm x \; \forall x \in \mathbb R}$ thus we are done

alibez wrote: fattypiggy123 wrote: Find all functions from the reals to the reals satisfying \[f(xf(y) + x) = xy + f(x)\] $f$ is injective ( This is obvious). now from $P(x,\frac{-f(x)}{x}) $ we have : $f(\frac{-f(x)}{x})=-1\: \: \forall x\neq 0$ so ($f$ is injective) $\frac{-f(x)}{x}=c\rightarrow f(x)=kx$ Solution is uncomplete becuase it doesnt work for all $k$, before sustitute your solution on the problem you get: $$xy=k^2xy \implies k=\pm 1$$Then $f(x)=\pm x$

Solution 1: Claim: $f$ is bijective Proof: Injectivity: Let $f(a)=f(b)$ $P(1,a)$ yields $f(f(a)+1)=a+f(1)$ $P(1,b)$ yields $f(f(b)+1)=b_f(1)$ Thus $a+f(1)=b+f(1), \Longrightarrow a=b$, therefore $f$ is injective $\square$. Surjectivity: $P(c,x)$ yields $f(cf(x)+c)=cx+f(c)$ thus $f$ is surjective $\square$. $P(f(0),0)$ yields $f(f(0)^2+f(0))=f(f(0))\overset{\text{from injectivity}}{\Longrightarrow} f(0)^2=0\Longrightarrow f(0)=0$ Since $f$ is surjective, there exists an $\alpha$ s.t. $f(\alpha)=-1$ $P(x,\alpha)$ yields $f(0)=\alpha x+f(x)\overset{\text{from injectivity}}{\Longrightarrow} f(x)=-\alpha x:=Q(x)$ $Q(\alpha)$ yields $\alpha^2=1\Longrightarrow \alpha=\pm 1$, thus $\boxed{f(x)=x, f(x)=-x, \forall x\in \mathbb{R}}$ $\blacksquare$ Solution 2: (The proof is similar, so we will be using the already known facts from the first proof) Note that if $f(x)=x$ is a solution, so is $f(x)=-x$ $P(f(x),-1)$ yields $f(f(x)f(-1)+f(x))=f(0)\overset{\text{from injectivity}}{\Longrightarrow} f(x)f(-1)+f(x)=0$ thus $f(-1)=-1, \forall x\in\mathbb{R}_{\neq 0}$ $P(x,-\frac{f(x)}{x})$ yields $f\bigg(x f\left(-\frac{f(x)}{x}\right)+x\bigg)=0\overset{\text{from injectivity}}{\Longrightarrow} xf\left(-\frac{f(x)}{x}\right)+x=0\Longrightarrow f\left(\frac{-f(x)}{x}\right)=f(-1)$, $\forall x\in \mathbb{R}_{\neq 0}$ Thus $\frac{-f(x)}{x}=-1\Longrightarrow f(x)=x,\forall x \in \mathbb{R}_{\neq 0}$, however since $f$ is surjective and $f(0)=0$, we can extend $f$ to all reals. So, $\boxed{f(x)=x, f(x)=-x, \forall x \in \mathbb{R}}$ $\blacksquare$

$P(1,y)$ gives $f(f(y)+1)=y+f(1)$ so $f$ is bijective. Let $f(c)=0$, $P(c,c)\Longrightarrow c^2=0$ so $f(0)=0$. $P(x,\frac{-f(x)}{x})\Longrightarrow f(xf(\frac{-f(x)}{x})+x)=0\Longrightarrow xf(\frac{-f(x)}{x})=-x\Longrightarrow f(\frac{-f(x)}{x})=-1$ by injectivity this gives $\frac{-f(x)}{x}=a$ where a is a constant, so $f$ is linear. Now plug in to get $f(x)=x$ or $f(x)=-x$.

Sol:- Let $P(x,y)$ be the assertion. $P(1,y) \Longrightarrow f(f(y)+1)=y+f(1)$ $P(x,y+f(1)) \Longrightarrow f(x(y+f(1)+1))=x(f(y)+1)+f(x)$ Substituting $y=-f(1)-1$ in above equation we get that $f$ is a linear polynomial. So let $f(x)=ax+b$.Substituting this in original equation we get that $a^2xy+abx=xy \implies a^2=1 ,ab=0 \implies a=\pm 1 , b=0$. So $f(x)=\pm x$. For the sake of contradiction $\exists r,s$ such that $f(r) \neq -r$ and $f(s) \neq s$. So $f(r)=r$ and $f(s)=-s$. $P(r,s) \Longrightarrow f(-rs+r)=rs+r$ Case $1$:- $f(-rs+r)=-rs+r$ We have $-rs+r=rs+r \implies rs=0 \implies $ either $r$ or $s$ equals $0$ contradicting the fact $f(r) \neq -r$ , $f(s) \neq s$. Case $2$:- $f(-rs+r)=rs-r$ We have $rs-r=rs+r \implies r=0$ contradicting the fact $f(r) \neq -r$. Hence only $2$ possible solutions are there $1.$:- $f(x)=x$ $2.$:- $f(x)=-x$ Substituting back in original equation we find that both of them indeed work