Statement:
There is only one distinct pair of numbers that multiply to a given number and sum to a given number.
Proof (almost complete):
$$x+y=c_1 {}{}{}{} (1)$$
$$xy=c_2 {}{}{}{} (2)$$
We have from (1) that $y=c_1-x$, hence:
$$x(c_1-x)=c_2$$
Solving for $x$ from the (1) we also have:
$$y(c_1-y)=c_2$$
These are the same equations and thus must have the same solution set $x=\{x_0,y_0\}$ and $y=\{x_0,y_0\}$. They are quadratics and thus have at most $2$ distinct solutions in $\mathbb{C}$, giving us two solutions to our system of equations $(x_0,y_0)$ and $(y_0,x_0)$, which can be treated as one set of solutions , if they have one we still have one set of solutions.
Above what is left to prove is why $(x_0,x_0)$ and $(y_0,y_0)$ can't be solutions in the case $x_0 \neq y_0$, and why the pairs $(x_0,y_0)$ and $(y_0,x_0)$ work (the later follows from the definition of $x_0$ and $y_0$. How can I do this?
Also, how can I prove this without the symmetry of the equations?
The reason I'm interested is that I'm trying to explain to myself why a factoring method works:
$$(dx+e)(fx+g)=dfx^2+efx+gdx+eg=ax^2+bx+c$$
Notice $(df)(eg)=(ef)(gd)=(a)(c)$ and $ef+gd=b$. If we rearrange $efx$ and $gdx$ around (switching spots) we still have a quadratic favorable by grouping. Now all that is left to figure out that this method of factoring quadratic works is to show that $ef$ and $gd$ are the only two elements that add to $df+gf=b$ and multiply to $efgd$, that is equivalent to the following question.
Thanks in advance.
