Is the Number 2 Truly Persistent?

[anemone]

Here is this week's POTW:

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This week POTW is a follow-up question from https://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-348-jan-8th-2019-a-25563.htmlProve that 2 is persistent.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

No one answered last week's problem. (Sadface)

But, you can check the suggested solution as follows:

Spoiler

Suppose $a+b+c+d=2$ and $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}=2$.

Consider the monic polynomial $f(x)$ with roots $a,\,b,\,c,\,d$. Let $f(x)$ have expansion

$f(x)=x^4-e_1x^3+e_2x^2-e_3x+e_4$

By Vieta's formulas, we have

$e_1=a+b+c+d=2$

$\dfrac{e_3}{e_4}=\dfrac{bcd+cda+dab+abc}{abcd}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}=2$

Thus, $f(x)$ has the form $x^4-2x^3+rx^2-2sx+s$.

Next, we consider the polynomial $g(x)=f(1-x)$, which is the monic polynomial with roots $1-a,\,1-b,\,1-c,\,1-d$. We have

$\begin{align*}g(x)&=(1-x)^4-2(1-x)^3+r(1-x)^2-2s(1-x)+2\\&=x^4-2x^3+rx^2+(2-2r+2s)x-(1-r+s)\end{align*}$

By Vieta's formulas again, we have

$\begin{align*}\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}+\dfrac{1}{1-d}&=\dfrac{-2(2-2r+2s)}{-(1-r+s)}\\&=2\end{align*}$

Therefore, 2 is persistent.