Find all functions `f:R\rightarrow R` such that: `f(x)f(y)=f(xy-1)+xf(y)+yf(x)` `\forall` `x,y` `\inR`

Find all functions `f: R \rightarrow R` Such that:
`f(x)f(y)=f(xy-1)+xf(y)+yf(x)`    `\forall` `x ,y``\inR`

Solution

Replace : `y=0` we will see that: `f(x).f(0)=f(-1)+x.f(0)`

Notice that: `f(0)` not equal to `0` then, `f(x)=x+c`     which `c` is a constant. We replaced to our hypothesis unsatisfactory. Then, `f(0)=0` from that we can see that: `f(-1)=0`.

From our hypothesis, we replace `x=y=1` then, `f(1)^2=2.f(1)` `\rightarrow` `f(1)=0` or `f(1)=2`.

If we replace `y=-1` into our hypothesis,

                    `f(-x-1)=f(x)` .

If we replace `y=-y-1` into our hypothesis, we will get:

    `f(x).f(-y-1)=f[x(-y-1)-1]+x.f(-y-1)+(-y-1)f(x)`

But, `f(-y-1)=f(y)` and `f[x(-y-1)-1]=f(xy+x)`

        `f(xy-1)+yf(x)+xf(y)=f(xy+x)+x.f(y)-(y+1)f(x)`        `\forall` `x ,y``\inR`

Then, `f(xy-1)+yf(x)=f(xy+x)+x.f(y)-(y+1)f(x)`                `\forall` `x ,y``\inR`

In here, `x` is not equal to `0` we replace `y=1/x` then

       `f(x+1)=(x+2)/x.f(x)` or `f(x-1)=(x-1)/(x+1).f(x)`

Right here, If we replace `y=1` into our first hypothesis, we will get:

        `f(x).f(1)=f(x-1)+x.f(1)+f(x)=(x-1)/(x+1).f(x)+x.f(1)+f(x)`

Case `f(1)=0` then, `f\equiv0`

Case `f(1)=2` then, `f(x)=x(x+1)`

Hence, there are two functions of `f(x)` which are :

                `f(x)=0`    and  `f(x)=x(x+1)`  for all `x\inR`

Solution by: Thin Sokkean