Prove that $x^2 + y^2 -1$ can't be a product of two linear equations How do I prove that $x^2 + y^2 -1$ can't be a product of two linear equations? 
I tried assuming that it is a product of $f(x,y) = ax + by +c$ and $g(x,y) = dx + ey +f$
and I finally found the following set of equations: 
$$ad = 1,$$
$$be = 1,$$
$$af + cd = 0,$$
$$bf + ce = 0,$$
$$cf = -1.$$
I should probably find a contradiction somewhere, but I am not sure how to solve this system and find this contradiction.
 A: I assume you multiplied and compared coefficients.
You should have six equations:
Coefficients of $x^2$ gives $ad=1$
Coefficients of $y^2$ gives $be=1$
Coefficients of $xy$ gives $ae+bd=0$
Coefficients of $x$ gives $af+cd=0$
Coefficients of $y$ gives $bf+ce=0$
Constant terms gives $cf=-1$
You missed out the coefficients of $xy$
Use just the top three equations: $ad=1$ means that $a$ and $d$ are both positive or both negative. $be=1$ means that $b$ and $e$ are both positive or both negative.
That means that $ae$ and $bd$ have the same sign as each other.
But $ae+bd=0$ means that they can't have the same sign as each other unless they are both equal to zero, and that is your contradiction.
A: Partial solution for the case of $\mathbb R$:
If $x^2 + y^2 -1= (ax + by +c)(dx + ey +f)$, then the equation
$$\tag{1}
x^2 + y^2 -1=0
$$
defines the union of two lines
$ax + by +c= 0$ and $dx + ey +f=0$. But we know that (1) defines a circle. We got a contradiction.
A: If $af=-cd$ and $ce=-bf$ then $acef=bcdf$, and $ae=bd$ because $cf=-1$. But $(ae)^2=ad\cdot be=1$, so $ae\ne0$. As @DavidMolano notes, we have one more equation from the $xy$ coefficient. This coefficient is $ae+bd=2ae$, rather than $0$ as required.
A: That means we've to prove that the given equation represents pair of straight lines. So, for a general second degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ to represent pair of straight line 
$$\Delta=\begin{vmatrix}
a&h&g\\
h&b&f\\
g&f&c
\end{vmatrix}=0$$
But here,$\Delta=-1\ne 0$. Hence the given expression can't be represented as product of two linear equations. Check this. 
