The function $f : \mathbb{R} \to \mathbb{R}$ satisfies $f(x) f(y) = f(x + y) + xy$ for all real numbers $x$ and $y.$ Find all possible functions $f.$ I was trying to solve the problem
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The function $f : \mathbb{R} \to \mathbb{R}$ satisfies
$f(x) f(y) = f(x + y) + xy$ for all real numbers $x$ and $y.$ Find all possible functions $f.$


I started by substituting in $0$ for both, to find that $f(0) = 1$, as it becomes $f(0)f(x) = f(x)$ which can only mean $f(0) = 1$. However past this point, I started struggling, as setting both to $1$, or $1$ and $0$ doesn't reveal anything new about the problem. Trying to set both $f(x)$ terms equal to each-other didn't help either. Thanks!
Here is a solution. $f(x)f(1)-x=f(x+1)$. So $f(-1)f(1)=0$ So either $f(1)=0$ or $f(-1)=0$.  Suppose $f(1)=0$. Then $f(x+1)=-x$. Hence $f(x)=-x+1$ for every $x$ which does  satisfy the condition. Assume $f(-1)=0$. Then $f(x-1) = x$ or $ f(x)=x+1$ which is Ok. Answer: $f(x)=x+1$.or $-x+1$.
As you said we can say f(0)=1 and then substituting y=-x, we get
$f(x)f(-x)=f(0)-x^2$
$f(x)f(-x)=1-x^2$
Considering polynomials, we can say that the function is a linear polynomial.
Assuming $f(x)=ax+b$ we get x+1, -x+1 as the possible functions.