mathisfun7 wrote: Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that \[f(xf(y)+y)+f(-f(x))=f(yf(x)-y)+y\] for all $x,y\in\mathbb{R}$ Let $P(x,y)$ be the assertion $f(xf(y)+y)+f(-f(x))=f(yf(x)-y)+y$ Let $a=f(0)$ 1) If $a=0$ : no solution ================ Suppose $a=0$ $P(x,0)$ $\implies$ $f(-f(x))=0$ $P(0,f(x))$ $\implies$ $f(f(x))=f(x)$ $P(-1,f(x))$ $\implies$ $f(x)=-f(f(x)(f(-1)-1))$ and so $f(LHS)=f(RHS)$ and so $f(x)=0$ $\forall x$, which is not a solution. Q.E.D. 2) If $a\ne -1$ : no solution =================== Suppose $a\ne 1$ 2.1) If $f(a-x)\ne a$, then $x\in f(\mathbb R)$ If $f(a-x)\ne a$, then : $P(\frac{a-x}{a-f(a-x)},0)$ $\implies$ $f(\frac{a(a-x)}{a-f(a-x)})+f(-f(\frac{a-x}{a-f(a-x)}))=a$ $P(\frac{a-x}{a-f(a-x)},a-x)$ $\implies$ $f(\frac{a(a-x)}{a-f(a-x)})+f(-f(\frac{a-x}{a-f(a-x)}))$ $=f(\text{something})+a-x$ Subtracting, we get $f(\text{something})=x$ Q.E.D. 2.2) $f(\frac 1{1-a})=a$ If $f(\frac 1{1-a})\ne a$ then (using 2.1) $\frac a{a-1}\in f(\mathbb R)$ and so $\exists t$ such that : $f(t)=\frac a{a-1}$ $P(0,0)$ $\implies$ $f(-a)=0$ $P(t,-a)$ $\implies$ $a=0$, impossible Q.E.D. 2.3) No such solution $f(\frac 1{1-a})=a$ $P(0,0)$ $\implies$ $f(-a)=0$ $P(\frac 1{1-a},0)$ $\implies$ $f(\frac a{1-a})=a$ $P(0,\frac a{1-a})$ $\implies$ $f(\frac a{1-a})=\frac a{1-a}$ And so $a=0$, impossible Q.E.D. 3) $f(x)=x+1$ $\forall x$ =============== We previously got $a=1$ and $f(-a)=f(-1)=0$ $P(0,x)$ $\implies$ $\boxed{f(x)=x+1}$ $\forall x$, which indeed is a solution Q.E.D.

If $f(\frac 1{1-a})\ne a$ then (using 2.1) $\frac a{a-1}\in f(\mathbb R)$ and so $\exists t$ such that : $f(t)=\frac a{a-1}$ in this part, something mistaken... $\frac 1{1-a}=\ a-x$ didn't mean $\frac a{a-1}\in f(\mathbb R)$

is there any better suloution??

Is there any other solution?

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Here's a cuter solution:) Let $P(x,y)$ denote the assertion of $f(xf(y)+y)+f(-f(x))=f(yf(x)-y)+y$. Claim. $f(0)=0$ or $f(0)=1$. Proof. $P(0,0)\implies f(-f(0))=0$. $P(0,x)\implies f(x)=x+f(xf(0)-x)$ and $P(-f(0),-f(0))\implies f(f(0))=2f(0)$. Thus, $f(f(0)^2-f(0))=f(0)=f(-f(0)+f(0)^2)$, therefore let $a=f(0)^2-f(0)$, meaning $f(a)=f(-a)=f(0)$. Now, $P(-f(0),a)\implies a=0\implies f(0)=0$ or $f(0)=1$. $\square$ If $f(0)=1$, then use $P(0,x)$ to obtain that $f(x)=x+1$ in this case. This obviously works. Now consider $f(0)=0$. We have $P(x,0)\implies f(-f(x))=0$, thus $P(0,f(x))\implies f(f(x))=f(x)$. Note that $f\not \equiv 0$, hence $\exists a$, so that $f(a)\neq 0$. Therefore, $$P\left(\frac{-a}{f(a)},a\right)\implies -f(af\left(\frac{-a}{f(a)}\right)-a)=a,$$furthermore applying $f$ to both sides, we get that $f(a)=0$, which contradicts $f(a)\neq 0$, no solutions here. We are done.

Let $P(x,y)$ denote the assertion $f(xf(y)+y)+f(-f(x))=f(yf(x)-y)+y$. The main part is finding $f(0)$. $P(0,0)\Rightarrow f(-f(0))=0$ $P(-f(0),-f(0))\Rightarrow f(f(0))=2f(0)$ $P(0,-f(0))\Rightarrow f(0)=f(f(0)-f(0)^2)$ $P(0,f(0))\Rightarrow f(0)=f(f(0)^2-f(0))$ $P(-f(0),f(0)^2-f(0))\Rightarrow f(0)\in\{0,1\}$ Case 1: $f(0)=0$ $P(x,0)\Rightarrow f(-f(x))=0$ If $f(x)\ne0$ for some $x$, we have: $P\left(-\frac x{f(x)},x\right)\Rightarrow -x\in f(\mathbb R)\Rightarrow f(x)=0$ and a contradiction. Then $f(x)=0$ doesn't work. Case 2: $f(0)=1$ $P(0,-1)\Rightarrow f(-1)=0$ $P(0,x)\Rightarrow\boxed{f(x)=x+1}$, which fits.

Alternate way to show $f(0)\in\{0,1\}$: Let $P(x,y)$ denote the assertion into the equation, $f(0)=c$ $P(0,0): f(-c)=0$ $P(-c,0): f(-c^2)=0$ $P(-c^2,-c^2): f(c^2)=c^2+c$ $P(-c,c^2): c=c^2 \Rightarrow c\in\{0,1\}$

Wow, there are a lot of users with a Snorlax avatar in this site.

Denote the assertion by $P(x,y).$

If $f(0)=1$ then $f(-1)=0$ and $P(0,x)$ yields $f(x)=x+1$ which clearly fits. Now consider $f(0)=0.$ Then $P(x,0)$ implies $f(-f(x))=0$ and $P(0,f(x))$ implies $f(f(x))=f(x).$ To conclude taking $f$ of $P(-1,f(x))$ yields $f(x)=f(f(x))=0$ and this fails.