Solving for x: Four Positive Real Numbers

[anemone]

Let $a,\,b,\,c,\,d$ be different positive real numbers such that $a+\dfrac{1}{b}=b+\dfrac{1}{c}=c+\dfrac{1}{d}=d+\dfrac{1}{a}=x$.

Find $x$.

 

[Albert1]

Let $a,\,b,\,c,\,d$ be different positive real numbers such that $a+\dfrac{1}{b}=b+\dfrac{1}{c}=c+\dfrac{1}{d}=d+\dfrac{1}{a}=x$.

Find $x$.

Spoiler

$4x=a+\dfrac{1}{a}+b+\dfrac{1}{b}+c+\dfrac{1}{c}+d+\dfrac{1}{d}$
using $AP\geq GP$
we have :$4x\geq 8$
$\therefore x\geq 2$

 

[anemone]

Spoiler

$4x=a+\dfrac{1}{a}+b+\dfrac{1}{b}+c+\dfrac{1}{c}+d+\dfrac{1}{d}$
using $AP\geq GP$
we have :$4x\geq 8$
$\therefore x\geq 2$

Thanks for participating, Albert...

Spoiler

But, I'm sorry, your answer that states that $x\geq 2$ isn't conclusive enough and it's wrong. Also, $x=2$ is certainly not a solution for the given original equation.:(

For your information, $x=\sqrt{2}$ is the answer to this problem.

 

[Albert1]

sorry I did not notice $a,b,c,d$ are different

 

[Opalg]

Let $a,\,b,\,c,\,d$ be different positive real numbers such that $a+\dfrac{1}{b}=b+\dfrac{1}{c}=c+\dfrac{1}{d}=d+\dfrac{1}{a}=x$.

Find $x$.

[sp]Let $f(s) = \dfrac1{x-s},$ and define a sequence of functions $f^{(n)}$ inductively by $f^{(n+1)}(s) = f\bigl(f^{(n)}(s)\bigr),$ with $f^{(1)}(s) = f(s).$ Then $\dfrac1{f(s)} = x-s,$ so that $s + \dfrac1{f(s)} = x.$ Thus if $s=a$ then $f(s) = b$, $f^{(2)}(s) = c$, $f^{(3)}(s) = d$ and $f^{(4)}(s) = a = s.$ So we are looking for points of order $4$ for $f$ (that is to say, fixed points for the function $f^{(4)}$).

Start with $f^{(2)}(s) = f\bigl(f(s)\bigr) = \dfrac1{x-\frac1{x-s}} = \dfrac{x-s}{x^2-1-sx}.$ Then $$f^{(4)}(s) = f^{(2)}\bigl(f^{(2)}(s)\bigr) = \frac{x - \frac{x-s}{x^2-1-sx}}{x^2-1 - \frac{x(x-s)}{x^2-1-sx}} = \frac{x^3 - sx^2 - 2x + s}{x^4 - sx^3 - 3x^2 + 2xs + 1}.$$ The condition $f^{(4)}(s) = s$ becomes $x^3 - sx^2 - 2x + s = s(x^4 - sx^3 - 3x^2 + 2xs + 1),$ which simplifies to $x(2-x^2)(s^2 - sx + 1) = 0.$

If $x=0$ then $f(s) = -\dfrac1s$, and $f^{(2)}(s) = s$, so that every point has order $2$. If $s^2 - sx + 1 = 0$ then $s = \dfrac1{x-s}$, so that $s$ is a fixed point of $f$, which is not what we want. So the only possibility to find points of order $4$ is if $2-x^2 = 0$, in other words $x = \pm\sqrt2$.

The question asks for all the numbers $a,b,c,d$ to be positive, so we must take $x=\sqrt2.$ The function $f(s) = \dfrac1{\sqrt2-s}$ satisfies $f^{(4)}(s) = s$ for all $s$. It also has the property $f(s) > s$ whenever $s<\sqrt2$; and $f(s)<0$ whenever $s>\sqrt2$. That means that it is never possible to choose all four numbers $a,b,c,d$ to be positive. So long as you allow them to take negative values, you can choose $a$ to be any real number except $0$, $\frac1{\sqrt2}$ or $\sqrt2$. Then the numbers $a$, $b = f(a)$, $c = f(b) = f^{(2)}(a)$ and $d = f( c) = f^{(3)}(a)$ will be distinct, and will satisfy the given equations. But one of those numbers will necessarily be negative: you can't have all four of them positive.[/sp]

 

[anemone]

[sp]Let $f(s) = \dfrac1{x-s},$ and define a sequence of functions $f^{(n)}$ inductively by $f^{(n+1)}(s) = f\bigl(f^{(n)}(s)\bigr),$ with $f^{(1)}(s) = f(s).$ Then $\dfrac1{f(s)} = x-s,$ so that $s + \dfrac1{f(s)} = x.$ Thus if $s=a$ then $f(s) = b$, $f^{(2)}(s) = c$, $f^{(3)}(s) = d$ and $f^{(4)}(s) = a = s.$ So we are looking for points of order $4$ for $f$ (that is to say, fixed points for the function $f^{(4)}$).

Start with $f^{(2)}(s) = f\bigl(f(s)\bigr) = \dfrac1{x-\frac1{x-s}} = \dfrac{x-s}{x^2-1-sx}.$ Then $$f^{(4)}(s) = f^{(2)}\bigl(f^{(2)}(s)\bigr) = \frac{x - \frac{x-s}{x^2-1-sx}}{x^2-1 - \frac{x(x-s)}{x^2-1-sx}} = \frac{x^3 - sx^2 - 2x + s}{x^4 - sx^3 - 3x^2 + 2xs + 1}.$$ The condition $f^{(4)}(s) = s$ becomes $x^3 - sx^2 - 2x + s = s(x^4 - sx^3 - 3x^2 + 2xs + 1),$ which simplifies to $x(2-x^2)(s^2 - sx + 1) = 0.$

If $x=0$ then $f(s) = -\dfrac1s$, and $f^{(2)}(s) = s$, so that every point has order $2$. If $s^2 - sx + 1 = 0$ then $s = \dfrac1{x-s}$, so that $s$ is a fixed point of $f$, which is not what we want. So the only possibility to find points of order $4$ is if $2-x^2 = 0$, in other words $x = \pm\sqrt2$.

The question asks for all the numbers $a,b,c,d$ to be positive, so we must take $x=\sqrt2.$ The function $f(s) = \dfrac1{\sqrt2-s}$ satisfies $f^{(4)}(s) = s$ for all $s$. It also has the property $f(s) > s$ whenever $s<\sqrt2$; and $f(s)<0$ whenever $s>\sqrt2$. That means that it is never possible to choose all four numbers $a,b,c,d$ to be positive. So long as you allow them to take negative values, you can choose $a$ to be any real number except $\sqrt2$. Then the numbers $a$, $b = f(a)$, $c = f(b) = f^{(2)}(a)$ and $d = f( c) = f^{(3)}(a)$ will be distinct, and will satisfy the given conditions. But you can't have all four of them positive.[/sp]

Awesome, Opalg! :D And thanks for participating!

Spoiler

Here is another method:

Expressing $a,\,b,\,c,\,d$ in terms of $x$ and $a$, we have

$d=\dfrac{ax-1}{a},\,c=\dfrac{ax^2-x-a}{ax-1},\,b=\dfrac{ax^3-x^2-2ax+1}{ax^2-x-a}$

Substitute the last expression into $a+\dfrac{1}{b}=x$, we have

$a+\dfrac{ax^2-x-a}{ax^3-x^2-2ax+1}=x$

$ax^4-a^2x^3-x^3-2ax^2+2a^2x+2x=0$

$x(x^2-2)(ax-a^2-1)=0$

Note that the last factor on the LHS of the last row is non-zero, for otherwise we get

$a+\dfrac{1}{a}=x=a+\dfrac{1}{b}$, which contradicts the fact that $a\ne b$. Since $x>0$, we must have $x=\sqrt{2}$.

Edit: Opalg is right, this problem is flawed since it is impossible to choose all four positive numbers that satisfy the given condition. Thanks, Opalg for this discovery.

 

[Albert1]

Let $a,\,b,\,c,\,d$ be different positive real numbers such that $a+\dfrac{1}{b}=b+\dfrac{1}{c}=c+\dfrac{1}{d}=d+\dfrac{1}{a}=x$.

Find $x$.

I have a question ,you said $x=\sqrt 2$
but by using $AP>GP$ (FOR $a,b,c,d$ are different and positive real numbers)
$4x=a+b+c+d+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}>8\sqrt[8]{abcd\times \dfrac{1}{abcd}}=8$
so $x>2>\sqrt 2$
how to explain it ? do I miss something ?

 

[anemone]

I have a question ,you said $x=\sqrt 2$
but by using $AP>GP$ (FOR $a,b,c,d$ are different and positive real numbers)
$4x=a+b+c+d+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}>8\sqrt[8]{abcd\times \dfrac{1}{abcd}}=8$
so $x>2>\sqrt 2$
how to explain it ? do I miss something ?

Hi again Albert!:)

Spoiler

Opalg has mentioned about the flaw of this problem in his solution and your intuition is right, this problem will work with three positive different real and the fourth will necessarily be a negative real.:)

 

[Albert1]

my solution:

Spoiler

let $b=\sqrt m+\sqrt {m-1}>0$
then $\dfrac {1}{b}=\sqrt m -\sqrt{m-1}>0$
for simplification let $b=\dfrac {1}{b}$
then $b=m=1---(1)$
and :$a=x-1>0 ---(2)$ so $x>1$
$c=\dfrac {1}{x-1}>0---(3)$
$d=\dfrac {x^2-x-1}{x-1}<0---(4)$
for $c+\dfrac {1}{d}=x$
we have :$\dfrac {1}{x-1}+\dfrac {x-1}{x^2-x-1}=x$
or $(x-1)^2=(x^2-x-1)^2$
we get :$x-1=\pm (x^2-x-1)$
and $x=0,2,\sqrt 2$
if $x=2$ then $a=b=c=d=1$
and $x=\sqrt 2$ is the solution we want