Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(xy - 1) + f(x)f(y) = 2xy - 1$ Using induction, I proved that $f(x) = x$ and $f(x) = -x^2$ work, but only for rational numbers.
How can I prove them for all real numbers?
 A: So we assume $$f(xy - 1) + f(x)f(y) = 2xy - 1\tag1$$ for all real $x,y$. The RHS is not constant, so constants aren't solutions. Substituting $y=0$ in (1) gives
$$f(-1)+f(x)f(0)=-1,$$ and as $f(x)$ is not a constant, we get
$$f(0)=0,\quad f(-1)=-1\tag2.$$
Letting $y=1$, we obtain $$f(x-1)+f(x)f(1)=2x-1\tag3.$$ Replacing $x$ by $xy$ in (3) gives $f(xy-1)+f(xy)f(1)=2xy-1$, and together with (1), this implies
$$f(x)f(y)=f(xy)f(1)\tag4.$$ But as the OP pointed out already in a comment, $f(1)$ can have only two possible values: from (3) with $x=1$, we get $f(0)+f(1)^2=1$, i.e. $f(1)=\pm1$.  
Case $f(1)=1$: From (3), we get $$f(x-1)=2x-1-f(x)\tag5,$$ from (4) with $y=-1$ we conclude $$f(-x)=f(x)f(-1)f(1)=-f(x)\tag{odd}.$$ Replacing $x$ by $-x$ in (5) gives $f(-x-1)=-2x-1-f(-x)$, using (odd), we arrive at
$$f(x+1)=2x+1-f(x)\tag6.$$
From (1) with $y=x$, we have $$f(x^2-1)+f(x)^2=2x^2-1\tag{7}.$$ But $x^2-1=(x-1)(x+1)$, so (using (5) and (6))
$$f(x^2-1)+f(x)^2=[2x-1-f(x)][2x+1-f(x)]+f(x)^2=2x^2-1,$$ and after some simplification, $$[f(x)-x]^2=0,$$ i.e. $f(x)=x$.  
Case $f(1)=-1$: Here, (4) with $y=-1$ gives
$$f(-x)=f(x)\tag{even},$$ and instead of (5) and (6), we get in the same way
$$f(x-1)=2x-1+f(x)$$ and $$f(x+1)=-2x-1+f(x).$$ Since in this case
$f(x^2-1)=-f(x-1)f(x+1)$, (7) becomes
$$-[2x-1+f(x)][-2x-1+f(x)]+f(x)^2=2x^2-1,$$ simplifying to
$f(x)=-x^2$.
