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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.

user1729
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NotanAI
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  • Where do your coefficients cone from? That is very crucial. – Allawonder Jun 09 '20 at 15:30
  • You are lacking an extra equation. Besides that, notice that the first two ones and the last one allow you to divide by $a,b,c,d,e,f$ whenever you want. – LeviathanTheEsper Jun 09 '20 at 15:34
  • Well, no matter where the coeffs lie, the two lines intersect in a point $P$. If $F(x,y)=x^2+y^2-1$, then you see that both $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial y}$ will be zero. But you see that both partials are never zero on the curve $F=0$. – Lubin Jun 09 '20 at 15:38
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    I removed the [group-theory] tag as I could not see why it was relevant. (If you think it is relevant and want to add it again, then maybe comment with the justification?) – user1729 Jun 09 '20 at 15:39

4 Answers4

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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.

tomi
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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.

AVK
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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.

J.G.
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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.

SarGe
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