Show that $\{a,b\}=\{c,d\}$ when $a+b+\sqrt{(a+b+1)^2-4ab}=c+d+\sqrt{(c+d+1)^2-4cd}$

Question

My approach is the following:

If $a+b=c+d$, then we obtain $ab=cd$ and the result follows.

If $a+b=c+d+p$, $p\ne 0$, then I obtain a complicated expression, from which I cannot deduce anything.

Any help is appreciated.

Answer 1

Your proposition is wrong.

Note that the condition is equivalent to these two parabolas having one root in common

$\begin{cases}x^2-(a+b+1)x+ab=0\\x^2-(c+d+1)x+cd=0\end{cases}$

Yet, it does not makes them identical, the other root can be different, affecting the coefficients.

For instance $f(6,10)=f(8,9)=23$

Gives parabolas $x^2-17x+60=(x-5)(x-12)\neq x^2-18x+72=(x-6)(x-12)$

$12$ is a common root but the other roots $5$ and $6$ differ, and the numbers $a,b,c,d$ are all different.