Suppose $x$ is a non zero real number such that both $x^5$ and $20x+\frac{19}{x}$ are rational numbers. Prove that $x$ is a rational number.
Problem
Source: RMO 2019 P1
Tags: RMO, algebra, rational numbers
20.10.2019 13:36
This same problem has already been posted a few seconds before yours: https://artofproblemsolving.com/community/c6h1937158_proving_rationality
20.10.2019 13:54
Consider the quadratic $20x^2-px+19=0$ with $p$ rational. Its roots are irrational(for contradiction).Write roots with quadratic and prove that it's 5th power cannot be rational using binomial.
20.10.2019 14:00
meet18 wrote: Consider the quadratic $20x^2-px+19=0$ with $p$ rational. Its roots are irrational(for contradiction).Write roots with quadratic and prove that it's 5th power cannot be rational using binomial. That's exactly what I did! I am getting a 17 on it
20.10.2019 14:29
Math-wiz wrote: meet18 wrote: Consider the quadratic $20x^2-px+19=0$ with $p$ rational. Its roots are irrational(for contradiction).Write roots with quadratic and prove that it's 5th power cannot be rational using binomial. That's exactly what I did! I am getting a 17 on it What was the motivation behind your solution.
20.10.2019 14:32
See, rational means $\frac{p}{q}$. All you could do is write first one as $\frac{p}{q}$ and play with 2nd, or vice versa. The second one seemed easier
20.10.2019 15:04
meet18 wrote: Consider the quadratic $20x^2-px+19=0$ with $p$ rational. Its roots are irrational(for contradiction).Write roots with quadratic and prove that it's 5th power cannot be rational using binomial. I was doing just that when decided it would be difficult to take the 5th power and there must be another way
20.10.2019 15:06
@above, sad life. Anyway, I hope you'll make it coz they're saying the paper was hard.
20.10.2019 15:08
MathPassionForever wrote: @above, sad life. Anyway, I hope you'll make it coz they're saying the paper was hard. How many did you solve, MPF? All?
20.10.2019 15:22
The solution to equation $20 x^2 - px + 19=0$ is of the form $a + b \sqrt c$ for some rational $a, b, c$. So $x^5 = (a+b\sqrt c)^5 = m+n\sqrt{c}$ where $m, n$ are some rationals and $n=5a^4b+10a^2b^3c+c^2$. If $n\neq 0$ then $\sqrt c$ must be rational for $x^5$ to be rational. If $n=0$, then $20 a^2=(-10\pm 2\sqrt{10}) b^2 c$ which is not possible for non-zero $b, c$. If $b$ or $c$ are $0$ then $x$ is clearly rational. Hence, we are done.
20.10.2019 15:25
See if my solution is correct 20x+19/x=m/n(given) 20x^5+19x^4=(m/n)x³ Now 20x^5 is rational so x³,x^4 is also rational so x^4/x³=x is also rational
20.10.2019 15:27
How can you conclude both $x^3, x^4$ rational? Irrational numbers are not closed under addition
20.10.2019 15:44
Math-wiz wrote: How can you conclude both $x^3, x^4$ rational? Irrational numbers are not closed under addition True, but if you can ensure that the irrationals you are adding are of $a + \sqrt{b}$ form($a,b \in \mathbb{Q} $) and that the irrational parts don't add up to $0$, then you can.
20.10.2019 16:00
My solution .... Given ,$20x+\frac{19}{x}$. Let, p is a rational so that, $p=20x+\frac{19}{x}$ Now ,let us multiply $x^5$ both side . Get,$px^5=20x^6+19x^4$. So,$px^5=r$ and $20x^5=k$. Where,$r,k $are rational. So, we are getting , $x(k+19x^3)=r$ Case (1): $x$ and$x^4$ both irrational . So we,are getting ,$r=0$. I.e ,$k=-19x^3$ since,$x\neq0$. but this giving$ x$imaginary so not possible . Case (2): $x$irrational but $x^4$ is rational So,we are getting,$r=19x^4$ and $k=0$ So, it is also not possible So,only one case left both are rationals. So .$x$ is rational.$\square$
20.10.2019 16:06
Ok, solution which I wrote, in summarised form $20x+\frac{19}{x}=\frac{p}{q}\implies 40x=\frac{p}{q}\pm\sqrt{D}$ Where $D$ is obviously rational. $40^5x^5=\left(\frac{p}{q}\pm\sqrt{D}\right)^5$ Hence, $\left(\frac{p}{q}\pm\sqrt{D}\right)^5$ is rational. Binomial expansion implies $\sqrt{D}$ is rational. Hence, $x$ is rational
20.10.2019 16:08
ftheftics wrote:
How did you get $r=0$ in case 1?
20.10.2019 16:09
Putting it here too for completeness Okay, so here is a brief sketch. Assume the contrary. If $20x +\frac{19}{x}$ is rational, then $x= a+b\sqrt{c}$ for $a,b$ rationals and $c$ a squarefree natural. But then $x^5$ has coefficient of $\sqrt{c}$ as $$5a^4b+10a^2b^3c + b^5c^2$$ this should be $0$, which gives $$a^2 = b^2c \cdot \frac{-10 \pm \sqrt{80}}{10}$$ a contradiction. There are tiny technicalities (zero, non-zero, etc.) that I have generously skipped.
20.10.2019 18:29
I did like this $20x^2-px+19=0$ $x^2=\frac{px-19}{20}$ ---(1) $x^3=\frac{px^2-19x}{20}$ Now the $x^2$ in the numerator can be replaced by equation Hence we obtain $x^3$ as a linear function of $x$ Similarly for $x^4$ and $x^5$ we get linear expressions And if $x^5=ax+b$ is rational and so are $a$ and $b$, we may conclude $x$ is rational I also considered the cases when denominator becomes zero, but am not including them here Please check my solution
20.10.2019 18:39
Arhaan wrote: ftheftics wrote:
How did you get $r=0$ in case 1? For equating rationals to irrational
20.10.2019 18:42
ftheftics wrote: Arhaan wrote: ftheftics wrote:
How did you get $r=0$ in case 1? For equating rationals to irrational But sum of 2 irrationals can be a rational.
21.10.2019 08:33
Redacted
21.10.2019 14:21
I don’t think this approach has been posted yet, so... First notice that $a^2x^2+ \frac{b^2}{x^2}$ is rational $=> (ax- \frac{b}{x})^2$ is rational, thus $ax-\frac bx =\sqrt{m}$, which when added to $ax+ \frac bx$ gives $x=k+\sqrt{l}$. As $x^5$ is rational, $\sqrt{l} (5k^2+10kl+l^2)$ is rational by the binomial expansion of $(k+\sqrt{l})^5$. Notice $5k^2+10kl+l^2=0 => l=0$, which gives $x=k$, thus we’re done, otherwise $\sqrt{l}$ is rational, which again gives $x=k+\sqrt{l}$ is rational. Thus we’re done.
21.10.2019 15:05
Math-wiz wrote: Ok, solution which I wrote, in summarised form $20x+\frac{19}{x}=\frac{p}{q}\implies 40x=\frac{p}{q}\pm\sqrt{D}$ Where $D$ is obviously rational. $40^5x^5=\left(\frac{p}{q}\pm\sqrt{D}\right)^5$ Hence, $\left(\frac{p}{q}\pm\sqrt{D}\right)^5$ is rational. Binomial expansion implies $\sqrt{D}$ is rational. Hence, $x$ is rational I just did the same thing 2 page long solution
21.10.2019 17:31
Assume that $20x+\frac{19}{x}=r\in\mathbb{Q}.$ Then $x^2=\frac{r}{20}x-\frac{19}{20}.$ Then $$x^3=\frac{r}{20}x^2-\frac{19}{20}x=\frac{r}{20}\left(\frac{r}{20}x-\frac{19}{20}\right)-\frac{19}{20}x=ax+b$$where $a,\,b\in\mathbb{Q}.$ In this way, $x^4$ and $x^5$ can also be written as a linear function of $x$ with rational coefficients. Suppose $x^5=a_5x+b_5$ where $a_5,\,b_5\in\mathbb{Q}.$ Since $x^5$ is rational, $x$ must be rational.
28.10.2019 06:49
I don't think $20$ and $19$ were of any use here. In fact it can be done for any rationals a,b
28.10.2019 07:18
@above ig its the year 2019 so they put it like that.
28.10.2019 11:59
The idea is as follows. It is given that $20x+\frac{19}{x}\in\mathbb{Q}$. It suffices to show $20x-\frac{19}{x}\in\mathbb{Q}$, because adding both will give us that $40x\in\mathbb{Q}$, which would imply $x\in\mathbb{Q}$. Since $x^5\in\mathbb{Q}$, we have $\frac{1}{x^5}\in\mathbb{Q}$. Also note that $(20x)^5$ and $\frac{19^5}{x^5}$ are both in $\mathbb{Q}$. Since $\left(20x+\frac{19}{x}\right)\in\mathbb{Q}$, we get $$\left(20x+\frac{19}{x}\right)^{2}=(20x)^2+\frac{19^2}{x^2} \in\mathbb{Q} \implies (20x)^4+\frac{19^4}{x^4}\in\mathbb{Q}$$Now we make use of the identity $$a^5-b^5 = (a-b)(a^4+a^3 b+ a^2 b^2 + ab^3 + b^4)$$where $a=20x$ and $b=\frac{19}{x}$. We have \begin{align*} (20x)^5 - \biggl(\frac{19}{x}\biggr)^{5} &= \left(20x-\frac{19}{x}\right) \cdot \left[ (20x)^4 + (20x)^3 \cdot \frac{19}{x} + (20\cdot 19)^2 + 20x \cdot \frac{19^3}{x^3} + \frac{19^4}{x^4}\right] \\ &= \left(20x-\frac{19}{x}\right) \cdot \left[\left\{(20x)^4+\frac{19^4}{x^4}\right\} + 20\cdot 19 \cdot \left\{(20x)^2+\frac{19^2}{x^2}\right\} + (20\cdot 19)^{2} \right] \end{align*}Each of the terms inside the curly brackets are in $\mathbb{Q}$ and the LHS is already in $\mathbb{Q}$, which forces $20x-\frac{19}{x}\in\mathbb{Q}$. \qed
28.10.2019 13:36
I tried to show same but could not prove it during exam. I did it till $(20x)^5- \left( \dfrac {19}{x} \right) ^5$. How much marks can I expect for this?
28.10.2019 15:17
DPS wrote: I tried to show same but could not prove it during exam. I did it till $(20x)^5- \left( \dfrac {19}{x} \right) ^5$. How much marks can I expect for this? You can expect 12+ hopefully.
13.11.2019 13:09
what if I do this? let 20x + 19/x = k, then form the quadratic. For rational roots, the condition would be that k is greater than or equal to 4 sqrt(95) then, if we use am gm inequality, we see that 20x + 19/x > or equal to 4 sqrt(95) so that means the quadratic has rational roots and hence x is rational. But this doesnt use the fact that x^5 is also rational PS. I didnt give the exam, I am just trying to solve the questions for fun
13.11.2019 13:21
TheLegendOfLegends wrote: what if I do this? let 20x + 19/x = k, then form the quadratic. For rational roots, the condition would be that k is greater than or equal to 4 sqrt(95) then, if we use am gm inequality, we see that 20x + 19/x > or equal to 4 sqrt(95) so that means the quadratic has rational roots and hence x is rational. But this doesnt use the fact that x^5 is also rational PS. I didnt give the exam, I am just trying to solve the questions for fun AM-GM won't work, terms can be negative. And, condition for rational roots is square root of discriminant should be rational
14.11.2019 11:31
oh, I messed up with the concepts. Thanks for the explanation
07.04.2020 07:22
I gave the perfect solution and they gave me 8/17 and even after reevaluation, the same. I am from tamilnadu. Can anybody tell who corrected the RMO there?
20.10.2020 15:49
ubermensch wrote: I don’t think this approach has been posted yet, so... First notice that $a^2x^2+ \frac{b^2}{x^2}$ is rational $=> (ax- \frac{b}{x})^2$ is rational, thus $ax-\frac bx =\sqrt{m}$, which when added to $ax+ \frac bx$ gives $x=k+\sqrt{l}$. As $x^5$ is rational, $\sqrt{l} (5k^2+10kl+l^2)$ is rational by the binomial expansion of $(k+\sqrt{l})^5$. Notice $5k^2+10kl+l^2=0 => l=0$, which gives $x=k$, thus we’re done, otherwise $\sqrt{l}$ is rational, which again gives $x=k+\sqrt{l}$ is rational. Thus we’re done. That's exactly what I did
06.11.2020 13:09
poplintos wrote: I gave the perfect solution and they gave me 8/17 and even after reevaluation, the same. I am from tamilnadu. Can anybody tell who corrected the RMO there? maybe you skipped some things about non-zero and zero most probably
08.08.2021 10:36
Jupiter_is_BIG wrote: Since, $x^5$ is rational, then so is $x^{10}\implies x^{10}\times\left(20x+\frac{19}{x}\right)=20(x^{10})\cdot x+19\cdot x^9$ and $ x^5\times\left(20x+\frac{19}{x}\right)=20(x^{5})\cdot x+19\cdot x^4$ are rational. Now, let $20x^5=t_1, 20x^10=t_2$ where $t_1,t_2$ are rational numbers. Thus, $t_1x+19x^4$ and $t_2x+19x^9$ are rational. Thus, $t_1t_2x+19t_2x^4$ and $t_2t_1x+19t_1x^9$ are rational and so is their difference. Thus, $19x^4(t_1x^5-t_2)$ is rational. Thus, $x^4$ is rational. Thus, $\frac{x^5}{x^4}=x$ is rational. Easiest solution,thanks jupiter.
03.03.2022 15:57
Proof. $20x + \frac{19}{x} \in \mathbb{Q} \Rightarrow 20x^2 + \frac{19}{x^2} \in \mathbb{Q} \Rightarrow 20x^3 + \frac{19}{x^3} \in\mathbb{Q} \Rightarrow 20x^6 + \frac{19}{x^6} \in \mathbb{Q}\Rightarrow (20x^5 + \frac{19}{x^5})(20x^6 + \frac{19}{x^6}) \Rightarrow 20x^{11} + \frac{19}{x^{11}} \in \mathbb{Q}$. $x^5(20x^6 +\frac{19}{x^6})\Rightarrow 20x^{11} + \frac{19}{x} \in \mathbb{Q} \Rightarrow 20x^{11} + \frac{19}{x} - (20x^{11} + \frac{19}{x^{11}}) \Rightarrow \frac{19}{x} - \frac{19}{x^{11}} \in \mathbb{Q} \Rightarrow (\frac{1}{x})( 1- \frac{1}{x^{10}}) \Rightarrow \frac{1}{x} \in \mathbb{Q} \Rightarrow x \in \mathbb{Q}$.$\blacksquare$
10.08.2023 14:20
Okay, sure the problem was easy. But if you are learning Algebraic Number Theory, or maybe just learning about minimial polynomials and roots of unity, then this problem is also an excellent training problem. Here's a solution by OMC YouTube Channel which features the use of minimal polynomials. For sure, I agree this is an overkill for such a problem, and much simpler solutions exist with just Binomial Theorem but this solution is worth giving a shot in my opinion. Solution: Let $\alpha$ be some fixed real number for which both of the given expressions are rational. Let \[20\alpha + \frac{19}{\alpha} = p \qquad \text{and} \qquad \alpha^5 = r\]for rational $p,r$. Consider the polynomials \begin{align*} P(x) &\coloneqq 20x^2 -px + 19 \\ Q(x) &\coloneqq x^5 - r. \end{align*}Then its not hard to verify that $\alpha$ is a root of both of the polynomials and therefore must be an algebraic number. Now, consider the minimal polynomial of $\alpha$ which we call $\pi_{\alpha}$. If $\deg(\pi_{\alpha})= 1$, then we're already done. Therefore assume for the sake of contradiction that $\deg(\pi_{\alpha}) = 2$. We now cite and prove a well-known theorem about minimal polynomials. Theorem. For any algebraic number $\alpha$, if $G(\alpha) = 0$ for $G \in \mathbb{Q}[x]$ then $\pi_{\alpha} \mid G(x)$. Proof: By the division algorithm, just write $G(x) = \pi_{\alpha}Q(x) + R(x)$ for $Q,R \in \mathbb{Q}[x]$ with $\deg(R) < \deg(\pi_{\alpha})$. Plugging $x = \alpha$ would give that $R(\alpha) = 0$ which a contradiction on minimality of degree unless $R \equiv 0$ which is as desired. $\square$ Applying the above theorem, we immediately get that $P(x)$ divides $Q(x)$. This means that roots of $P$ are a subset of roots of $Q$. By roots of unity, one can easily find out the roots of $Q$. Setting $\omega = e^{\frac{2i\pi}{5}}$, the set of roots of $Q$ is precisely \[R_Q = \{x \colon x = \sqrt[5]{r}\cdot\omega^i \,\, \text{for $1 \le i \le 5$ and $i \in \mathbb{Z}$}\}.\]Essentially, we now need to find two elements of $R_Q$ such that both their sum and product is rational so that we can get the roots of $P(x)$. This occurs precisely when you take the complex conjugate pairs in $R_Q$. If you pick $1$ as the common root, then it follows quickly $1$ is a double root of $P$. But this is impossible since it would get us $p = 39$ with $P(1) = 0$ but $p = 40$ with $P'(1) = 0$. There are two more cases when you pick $\{\sqrt[5]{r}\omega, \sqrt[5]{r}\omega^4\}$ or $\{\sqrt[5]{r}\omega^2, \sqrt[5]{r}\omega^3\}$. We only consider the first one since other one follows symmetrically. Write \[P(x) = 20x^2 - px + 19 = 20(x-\sqrt[5]{r}\omega)(x-\sqrt[5]{r}\omega^4).\]Comparing constant terms, we would immediately get a contradiction. We ditch this approach, since we can prove something lot stronger i.e., we can replace $20,19$ by any rational constants and the result holds true. From comparing constant terms, $m \coloneqq \sqrt[5]{r^2} \in \mathbb{Q}$. From coefficient of linear terms, we would get $\sqrt{m}\cos\left( \frac{\pi}{5} \right) \in \mathbb{Q}$. From here on, you just put $\cos\left(\frac{\pi}{5}\right) = \frac{\sqrt{5}+1}4$ and solve to eventually get a contradiction on the fact that $m$ is rational. This means our original assumption that $\deg(\pi_{\alpha}) > 1$ was wrong, therefore, $\alpha$ must be rational what had to proven. $\blacksquare$
10.10.2024 19:50
This is not the best method but It is easy. Since $x^5$ is rational number let it be equal to k. $x=k^{\frac{1}{5}}$ So for contradiction let x be an irrational number i.e. k can't be written with power which is an integral multiple of 5. Now, let $20x +\frac{19}{x} = q$ where q is a rational number. $20x^2-qx+19=0$ Let the two values of x satisying the equation be $x_1 and x_2 $ So $x_1+x_2=\frac{q}{20} $ which is a rational number. Since the equation has rational coefiencts and irrational roots. the roots must be conjugate of each other. So let $x_1= a+ k^{\frac{1}{5}}$ and$ x_2=a- k^{\frac{1}{5}}$ Also $ x_1x_2 = (a+ k^{\frac{1}{5}}) (a- k^{\frac{1}{5}})=\frac{19}{20}$ $a^2-k^\frac{2}{5}=\frac{19}{20}$ which is not possible since k can't have a power which is multiple of 5 by our assumption. This contradics our assumption and hence x is a rational number.
18.01.2025 21:01
It is given that $20x+\frac{19}{x}\in \mathbb{Q}$ and $x^5\in \mathbb{Q}$. We show that $20x-\frac{19}{x}\in \mathbb{Q}$. Note that \[(20x)^2+\frac{19^2}{x^2}\in \mathbb{Q}\implies (20x)^4+\frac{19^4}{x^4}\in \mathbb{Q}\]Now $\underbrace{(20x)^5-\left(\frac{19}{x}\right)^5}_{\in \mathbb{Q}}=\left(20x-\frac{19}{x}\right)\left(\underbrace{(20x)^4+\left(\frac{19}{x}\right)^4}_{\in \mathbb{Q}}+280\underbrace{\left((20x)^2+\frac{19^2}{x^2}\right)}_{\in \mathbb{Q}}+380^2\right)$ So $20x-\frac{19}{x}\in \mathbb{Q}$ and we get $40x\in \mathbb{Q}\implies x\in \mathbb{Q}$.