nAalniaOMliO wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for every $x,y \in \mathbb{R}$ the following equation holds:$$1+f(xy)=f(x+f(y))+(y-1)f(x-1)$$ I think it's not hard. Step 1: $f bijective$ Step 2: Using $P(x,1)$ we done.

nAalniaOMliO wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for every $x,y \in \mathbb{R}$ the following equation holds:$$1+f(xy)=f(x+f(y))+(y-1)f(x-1)$$ Let $P(x,y)$ be the assertion $1+f(xy)=f(x+f(y))+(y-1)f(x-1)$ Let $a=f(0)$ and $c=f(1)$ $P(x,1)$ $\implies$ $f(x+c)=f(x)+1$ and so $c\ne 0$ So, subtracting $P(0,\frac xc)$ from $P(c,\frac xc)$, we get $f(x)=\frac xc+a$ Plugging this back in original equation, we get three solutions : $\boxed{\text{S1 : }f(x)=x\quad\forall x\in\mathbb R}$, which indeed fits $\boxed{\text{S2 : }f(x)=\frac{1+\sqrt 5}2x-1\quad\forall x\in\mathbb R}$, which indeed fits $\boxed{\text{S3 : }f(x)=\frac{1-\sqrt 5}2x-1\quad\forall x\in\mathbb R}$, which indeed fits

$P(x, 1)$ gives $1+f(x)=f(x+f(1))$ $P(0, x)$ gives $1+f(0)=f(f(x))+(y-1)f(-1)$ so $f(f(x))$ is linear. Let $f(f(x))=ax+b$. First we deal with the case $a\neq 0$. $P(1, f(x))$ gives $1+ax+b=f(1+ax+b)+f(x)f(0)-f(0)$ $P(f(x), 1)$ gives $1+ax+b=f(f(x)+f(1))$ So $f(1+ax+b)=f(f(f(x)+f(1)))=a(f(x)+f(1))+b$ Therefore, $1+ax+b=a(f(x)+f(1))+b+f(x)f(0)-f(0)$ which means $f$ is linear. Let $f(x)=cx+d$, we have $1+cxy+d=cx+c^2y+cd+d+(y-1)(cx-c+d)$. Solving yields $c=\frac{\sqrt{5}+1}{2}$ or $c=\frac{-\sqrt{5}+1}{2}$ and $d=-1$. or $c=1$, $d=0$. If $a=0$ ($f(f(x))$ is constant. Then $1+f(0)=b+(y-1)f(-1)$, so $f(-1)=0$ Also $P(1, f(x))$ gives $1+b=f(1+b)+(f(x)-1)f(0)$. So either $f(x)=1$ (does not work) or $f(0)=0$. Now $P(x, 0)$ gives $1=f(x)-f(x-1)$, but $f(0)-f(-1)\neq 1$ so contradiction. Therefore the solutions are $f(x)=\frac{\sqrt5+1}{2}x-1$, $f(x)=\frac{-\sqrt5+1}{2}-1$ and $f(x)=x$.