This same problem has already been posted a few seconds before yours: https://artofproblemsolving.com/community/c6h1937158_proving_rationality

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.

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

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.

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

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

@above, sad life. Anyway, I hope you'll make it coz they're saying the paper was hard.

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?

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.

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

How can you conclude both $x^3, x^4$ rational? Irrational numbers are not closed under addition

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.

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.$\square$

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

ftheftics wrote:

How did you get $r=0$ in case 1?

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.

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

Arhaan wrote: ftheftics wrote:

How did you get $r=0$ in case 1? For equating rationals to irrational

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.

Redacted

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.

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

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.

I don't think $20$ and $19$ were of any use here. In fact it can be done for any rationals a,b

@above ig its the year 2019 so they put it like that.

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

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?

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.

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

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

oh, I messed up with the concepts. Thanks for the explanation

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?

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

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

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.

Proof. $20x + \frac{19}{x} \in \mathbb{Q} \Rightarrow 20x^2 + \frac{19}{x^2} \in \mathbb{Q} \Rightarrow 20x^3 + \frac{19}{x^3} \in\mathbb{Q} \Rightarrow 20x^6 + \frac{19}{x^6} \in \mathbb{Q}\Rightarrow (20x^5 + \frac{19}{x^5})(20x^6 + \frac{19}{x^6}) \Rightarrow 20x^{11} + \frac{19}{x^{11}} \in \mathbb{Q}$. $x^5(20x^6 +\frac{19}{x^6})\Rightarrow 20x^{11} + \frac{19}{x} \in \mathbb{Q} \Rightarrow 20x^{11} + \frac{19}{x} - (20x^{11} + \frac{19}{x^{11}}) \Rightarrow \frac{19}{x} - \frac{19}{x^{11}} \in \mathbb{Q} \Rightarrow (\frac{1}{x})( 1- \frac{1}{x^{10}}) \Rightarrow \frac{1}{x} \in \mathbb{Q} \Rightarrow x \in \mathbb{Q}$.$\blacksquare$

Problem Defence is mediocre

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. $\blacksquare$

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.

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 $40x\in \mathbb{Q}\implies x\in \mathbb{Q}$.