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.
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.
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 
$$\Delta=\begin{vmatrix}
a&amp;h&amp;g\\
h&amp;b&amp;f\\
g&amp;f&amp;c
\end{vmatrix}=0$$

But here,$\Delta=-1\ne 0$. Hence the given expression can't be represented as product of two linear equations. Check this.