Surds - - - challenge problem

[Bob Brown MSEE]

This is a problem inspired by Lookagain's solution to "Surd-Equation-Please-help"

Golden Surd Challenge #1:

Let a = \(\displaystyle 5 + 8{\phi }\), and b = \(\displaystyle 5 - \frac{8}{\phi }\)
use the defined Golden Ratio Constant \(\displaystyle {\phi }\) = 1.618...

Solve for x and y (rational numbers):

x + y\(\displaystyle \sqrt{5}\) = \(\displaystyle \sqrt[3]{a}\text{ + }\sqrt[3]{b}\)

(provide exact answers)

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Lookagain is likely to get this right, but everyone should try it. I think you will find it fun!
The result is quite unexpected!

 

[soroban]

Hello, Bob Brown MSEE!

\(\displaystyle \text{Let }a \,=\, 5 + 8\phi\,\text{ and }\,b \,=\, 5 - \frac{8}{\phi},\:\text{ where }\phi\text{ is the }Golden\:Ra{t}io\!:\:\frac{1+\sqrt{5}}{2}\)

\(\displaystyle \text{Solve for }x\text{ and }y\!:\;\sqrt[3]{a} + \sqrt[3]{b} \;=\;x + y\sqrt{5}\)

I used some fascinating facts about \(\displaystyle \phi\) to crank out these powers.

\(\displaystyle \begin{array}{ccc}\phi &=& \phi \\ \phi^2 &=& \phi + 1 \\ \phi^3 &=& 2\phi+1 \\ \phi^4 &=& 3\phi + 2 \\ \phi^5 &=& 5\phi+3 \\ \phi^6 &=& 8\phi+5 \end{array} \qquad\qquad \begin{array}{ccc}\frac{1}{\phi} &=& \phi - 1 \\ \frac{1}{\phi^2} &=& \text{-}(\phi-2) \\ \frac{1}{\phi^3} &=& 2\phi-3 \\ \frac{1}{\phi^4} &=& \text{-}(3\phi-5) \\ \frac{1}{\phi^5} &=& 5\phi-8 \\ \frac{1}{\phi^6} &=& \text{-}(8\phi-13)\end{array}\)


\(\displaystyle a \:=\:8\phi + 5 \:=\:\phi^6 \quad\Rightarrow\quad \sqrt[3]{a} \:=\:\phi^2 \:=\:\phi + 1\)

\(\displaystyle b \:=\:\dfrac{5\phi-8}{\phi} \:=\:\dfrac{\frac{1}{\phi^5}}{\phi} \:=\:\frac{1}{\phi^6} \quad\Rightarrow\quad \sqrt[3]{b} \:=\:\frac{1}{\phi^2} \:=\: -(\phi - 2) \)


Therefore: .\(\displaystyle \sqrt[3]{a} + \sqrt[3]{b} \;=\;(\phi + 1) - (\phi - 2) \;=\;\color{purple}{3}\)

 

[Bob Brown MSEE]

Hello, Bob Brown MSEE!

I used some fascinating facts about \(\displaystyle \phi\) to crank out these powers.

Therefore: .\(\displaystyle \sqrt[3]{a} + \sqrt[3]{b} \;=\;(\phi + 1) - (\phi - 2) \;=\;\color{purple}{3}\)

Yes 100%,
soroban You Win! VERY clever.
excellent!! (x, y) = (3, 0)

 

[Bob Brown MSEE]

Surd Challenge #2:

Let a = 40 + 11 \(\displaystyle \sqrt{13}\), and b = 40 - 11 \(\displaystyle \sqrt{13}\)

Solve for x and y (rational numbers):

x + y\(\displaystyle \sqrt{13}\) = \(\displaystyle \sqrt[3]{a}\text{ + }\sqrt[3]{b}\)

(provide exact answers)

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soroban is likely to get this right, but everyone should try it. I think you will find it fun!

 

[soroban]

Hello, Bob Brown MSEE!

We don't have a [hide] or

Spoiler

command.
Any suggestions?

Spoiler

\(\displaystyle \text{Let }\,a\,=\,40 + 11\sqrt{13}\:\text{ and }\:b \,=\, 40 - 11\sqrt{13}\)

\(\displaystyle \text{Solve for }x\text{ and }y\!:\;x + y\sqrt{13}\:=\: \sqrt[3]{a} +\sqrt[3]{b}\)

Let \(\displaystyle X \:=\:\sqrt[3]{a} + \sqrt[3]{b}\)

Cube both sides: .\(\displaystyle X^3 \;=\;\left(\sqrt[3]{a} + \sqrt[3]{b}\right)^3\)

. . \(\displaystyle X^3 \;=\;a + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} + b \)

. . \(\displaystyle X^3 \;=\;3\sqrt[3]{ab}\left(\sqrt[3]{a} + \sqrt[3]{b}\right) + (a+b)\)


Note that: .\(\displaystyle ab \:=\: (40 + 11\sqrt{13})(40-11\sqrt{13}) \:=\:1600 - 1573 \:=\:27\)
. . Hence: .\(\displaystyle \sqrt[3]{ab} \:=\:3\)

Also that: .\(\displaystyle a + b \:=\: (40+11\sqrt{13}) + (40-11\sqrt{13}) \:=\:80\)


The equation becomes: .\(\displaystyle X^3 \:=\:9\underbrace{(\sqrt[3]{a} + \sqrt[3]{b})}_{\text{This is }X} + 80 \)

We have: .\(\displaystyle X^3 \:=\: 9X + 80 \quad\Rightarrow\quad X^3 - 9X - 80 \:=\:0\)

And the only real root is \(\displaystyle X = 5.\)

Therefore: .\(\displaystyle \sqrt[3]{a} + \sqrt[3]{b} \;=\;5 \quad\Rightarrow\quad \begin{Bmatrix}x \:=\:5 \\ y\:=\:0\end{Bmatrix}
\)

 

[Bob Brown MSEE]

Hello, Bob Brown MSEE!

We don't have a [hide] or

Spoiler

command.
Any suggestions?

And the only real root is \(\displaystyle X = 5.\)


Therefore: .\(\displaystyle \sqrt[3]{a} + \sqrt[3]{b} \;=\;5\)

Spoiler

soroban,

Haha, we do need a spoiler command with you!!!

Congrats again.
(x,y) = (5,0) is the solution.
This proof is the approach I used for Challenge #1 too.
However I very much preferred your use of the
Golden ratio recursive power derivations to my solution on #1

Brilliant.

 

[soroban]

If you aren't familiar with the "fascinating facts" about \(\displaystyle \phi\), here they are.


The Golden Ratio arises from the equation: .\(\displaystyle x^2 - x - 1 \:=\:0\)
The positive root is: .\(\displaystyle x \:=\:\frac{1+\sqrt{5}}{2} \:=\:\phi\)

So we have: .\(\displaystyle \phi^2 - \phi - 1 \:=\:0 \quad\Rightarrow\quad \boxed{\phi^2 \:=\:\phi + 1}\)

We will use this identity to compute higher powers of \(\displaystyle \phi.\)


\(\displaystyle \begin{array}{cccccccccccc}\phi^3 &=& \phi\cdot\phi^2 &=& \phi(\phi+1) &=& \phi^2 + \phi &=& (\phi+1) + \phi &=& 2\phi + 1 \\ \phi^4 &=& \phi\cdot\phi^3 &=& \phi(2\phi+1) &=& 2\phi^2+\phi &=& 2(\phi + 1) + \phi &=& 3\phi + 2 \\ \phi^5 &=& \phi\cdot\phi^4 &=& \phi(3\phi+2) &=& 3\phi^2+2\phi &=& 3(\phi + 1) + 2\phi &=& 5\phi + 3 \\ \phi^6 &=& \phi\cdot\phi^5 &=& \phi(5\phi+3) &=& 5\phi^2+3\phi &=& 5(\phi+1)+3\phi &=& 8\phi + 5 \end{array}\)


Note that the coefficients are Fibonacci numbers: 1, 1, 2, 3, 5, 8, 13, . . .

. . In general: .\(\displaystyle \phi^n \;=\;(F_n)\phi + (F_{n-1})\)