Find all the functions $ f\left( x+y\right) +xy=f\left( x\right) f\left( y\right) $ Find all the functions $f:[0,\infty )\rightarrow R, f\left( x+y\right) +xy=f\left( x\right) f\left( y\right) , \forall x,y\in [0,\infty) $ 
I found $f(x)=0, if f(0)=0$,  and I proved  that $ f(n)=n+1 , f(n)=n-1, \forall n\in N$ are both solutions if $f(0)=1$ .
But how can I get the general solution?
 A: Set $x \to x+1$ and $y=1$:
$$f(x+2)=f(x+1)f(1)-(x+1)=[f(x)f(1)-x]f(1)-(x+1)$$
$$=f^2(1)f(x)-xf(1)-(x+1)$$
Set $x = 2$ and $y\to x$:
$$f(2+x)=f(2)f(x)-2x=(f^2(1)-1)f(x)-2x$$
Subtracting the two equations:
$$f(x)=(f(1)-1)x+1$$
Now set $y\to x$ and replace $f(x)=(f(1)-1)x+1$ to get the two solutions $f(x)=1+x$ and $f(x)=1-x$ which both check.
A: There is a linked post which deals with a similar functional equation, that gets closed, I'll show how they are related and also propose an alternate solving.
Let substitute $f(x)=-g(x)$
Notice we have $$f(x+y)+xy=f(x)f(y)\iff g(x)g(y)+g(x+y)=xy$$

Let now solve the problem for $g$:

*

*We have $g(x)g(0)+g(x)=0$
If $g(0)\neq -1$ then $\forall x\in\mathbb R,\ g(x)=0$ which is clearly not a solution since $g(1)g(1)+g(1)=1$ is not satisfied.

*

*Therefore $g(0)=-1$ then

$g(x)g(-x)+g(0)=-x^2\iff g(x)g(-x)=1-x^2$
In particular $g(1)g(-1)=0$

*

*case $g(1)=0$
$g(x-1)g(1)+g(x)=x-1\iff g(x)=x-1$
And we can verify it is a suitable solution
$\require{cancel}g(x)g(y)+g(x+y)=(x-1)(y-1)+(x+y-1)=(xy-\cancel{x}-\cancel{y}+\cancel{1})+(\cancel{x}+\cancel{y}-\cancel{1})=xy$

*

*case $g(-1)=0$
$g(x+1)g(-1)+g(x)=-(x+1)\iff g(x)=-(x+1)$
And we can verify it is a suitable solution
$\require{cancel}g(x)g(y)+g(x+y)=(x+1)(y+1)-(x+y+1)=(xy+\cancel{x}+\cancel{y}+\cancel{1})-(\cancel{x}+\cancel{y}+\cancel{1})=xy$
Conclusion

$$g(x)=-1\pm x\quad \text{or equivalently}\quad f(x)=1\mp x$$

