Let $x$ be a non-zero real numbers such that $x^4+\frac{1}{x^4}$ and $x^5+\frac{1}{x^5}$ are both rational numbers. Prove that $x+\frac{1}{x}$ is a rational number.
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pl210741
14.12.2013 00:05
Let $x+\frac{1}{x}=a$. Then we have:
\begin{align*}
x^2+\frac{1}{x^2}&=\left(x+\frac{1}{x}\right)^2-2=a^2-2\\
x^3+\frac{1}{x^3}&=\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)-\left(x+\frac{1}{x}\right)=a^3-3a\\
x^4+\frac{1}{x^4}&=\left(x^2+\frac{1}{x^2}\right)^2-2=a^4-4a^2+2\\
x^5+\frac{1}{x^5}&=\left(x^4+\frac{1}{x^4}\right)\left(x+\frac{1}{x}\right)-\left(x^3+\frac{1}{x^3}\right)=a^5-5a^3+5a.
\end{align*}
Now, let $P=a^4-4a^2+2$ and $Q=a^5-5a^3+5a$. Because $P$ and $Q$ are both rational, we know that $P+Q$ is also rational. But
\[P+Q=a^5+a^4-5a^3-4a^2+5a+2=(a-1)(a+2)(a^3-3a+1),\]
so if $a$ were irrational, $P+Q$ would be as well; contradiction. It follows that $a$ is rational.
ssilwa
14.12.2013 01:48
If $P$ and $Q$ are rational, then so is $Q/(P-2)$. Doing so gives us $a+1/(8a-16)+1/(8a+16)-5/4a$ which is rational. Maybe it is easier to deduce from here weather $a$ has to be rational?
kapilpavase
14.12.2013 09:28
it is easier if you solve it this way $ \left(x^4 +\dfrac {1}{x^4}\right)^2 -2=x^8+\dfrac{1}{x^8}$ a rational number similarly $ x^{10}+\dfrac{1}{x^{10}}$ is rational and ${\left(x^4 +\dfrac {1}{x^4}\right)^3-3\left(x^4 +\dfrac {1}{x^4}\right)=x^{12}}+\dfrac{1}{x^{12}}$is rational we have${ \left(x^{10}+\dfrac{1}{x^{10}}\right)\left(x^2+\dfrac{1}{x^2}\right)=x^8+\dfrac{1}{x^8}+x^{12}}+\dfrac{1}{x^{12}}$ which implies $ x^2+\dfrac{1}{x^2}$ is also rational .this in turn implies $x^6+\dfrac{1}{x^6}$to be a rational number. we again have $\left(x^5+\dfrac{1}{x^5}\right)\left(x+\dfrac{1}{x}\right)=x^6+\dfrac{1}{x^6}+x^4 +\dfrac {1}{x^4}$and it can be seen that $x+\dfrac{1}{x}$ is a rational number.
fmasroor
15.12.2013 00:12
Haha very clever kapilpavase! I like your solution a lot although it does seem a bit unmotivated.
MSTang
24.07.2014 08:26
pl210741 wrote:
Let $x+\frac{1}{x}=a$. Then we have:
\begin{align*}
x^2+\frac{1}{x^2}&=\left(x+\frac{1}{x}\right)^2-2=a^2-2\\
x^3+\frac{1}{x^3}&=\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)-\left(x+\frac{1}{x}\right)=a^3-3a\\
x^4+\frac{1}{x^4}&=\left(x^2+\frac{1}{x^2}\right)^2-2=a^4-4a^2+2\\
x^5+\frac{1}{x^5}&=\left(x^4+\frac{1}{x^4}\right)\left(x+\frac{1}{x}\right)-\left(x^3+\frac{1}{x^3}\right)=a^5-5a^3+5a.
\end{align*}
Now, let $P=a^4-4a^2+2$ and $Q=a^5-5a^3+5a$. Because $P$ and $Q$ are both rational, we know that $P+Q$ is also rational. But
\[P+Q=a^5+a^4-5a^3-4a^2+5a+2=(a-1)(a+2)(a^3-3a+1),\]
so if $a$ were irrational, $P+Q$ would be as well; contradiction. It follows that $a$ is rational.
Why is your last sentence true? If $a$ is irrational, I don't see a clear reason that the product $(a-1)(a+2)(a^3-3a+1)$ also has to be irrational.
fmasroor
26.07.2014 05:32
Yeah rereading this thread gave me that issue; take a= real root of that last factor
utkarshgupta
30.07.2014 14:06
Here is the generalization http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=6325&p=624252#p624252
aayush-srivastava
22.11.2015 20:47
At first I felt that this is similar to BrMO R1(P4).. But thery are different..But there's a solution using Vieta Jumping as well(All thnx. to $anantmudgal09$ for explaining it to me)
math_and_me
05.11.2019 04:48
Can you elaborate your solution (the vieta jumping one..)
HoRI_DA_GRe8
13.08.2021 17:46
math_and_me wrote: Can you elaborate your solution (the vieta jumping one..) Cant we take $(x^4+1/x^)(x+1/x-1)=x^5+1/x^5$ ???
Gryphos
13.08.2021 18:33
nathantareep wrote: math_and_me wrote: Can you elaborate your solution (the vieta jumping one..) Cant we take $(x^4+1/x^)(x+1/x-1)=x^5+1/x^5$ ??? I suppose you mean $(x^4+1/x^4)(x+1/x-1)=x^5+1/x^5$. But this is just wrong!!! In fact, $$\left( x^4 + \frac{1}{x^4} \right) \left( x + \frac{1}{x} - 1 \right) = x^5 + \frac{1}{x^5} - x^4 - \frac{1}{x^4} + x^3 + \frac{1}{x^3}.$$So you would need to show that $x^3 + x^{-3}$ is also rational...