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Source: 2016 Taiwan TST Round 3

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115 posts

#1 h Jul 25, 2016, 3:54 PM • 8 Y

Y by v_Enhance, 62861, anantmudgal09, freefrom_i, Davi-8191, Aryan-23, Adventure10, Mango247

Let $\lambda$ be a positive real number satisfying $\lambda=\lambda^{2/3}+1$ . Show that there exists a positive integer $M$ such that $|M-\lambda^{300}|<4^{-100}$ .

Proposed by Evan Chen

This post has been edited 1 time. Last edited by v_Enhance, Mar 9, 2017, 5:37 PM
Reason: claim authorship

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GoJensenOrGoHome

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GoJensenOrGoHome

#3 h Jul 25, 2016, 10:41 PM • 2 Y

Y by Adventure10, Mango247

This really looks like Dirichlet's approximation theorem

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Bajisu

#4 h Aug 7, 2016, 8:33 AM • 2 Y

Y by Adventure10, Mango247

weak dirichlet

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kk108

#5 h Aug 7, 2016, 9:08 AM • 1 Y

Y by Adventure10

Dirichlet's approximation theorem ?

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688 posts

#6 h Feb 25, 2017, 2:23 PM • 4 Y

Y by GoJensenOrGoHome, shinichiman, Adventure10, Mango247

Let $\alpha,\beta,\gamma$ be 3 complex roots of $t^3-t^2-1=0$ . And $\alpha$ is real, the other two is conjugate complex numbers.
Then $\lambda=\alpha^{3}$ . So $\lambda^{300}=\alpha^{900}$ .
Note that $\alpha^{900}+\beta^{900}+\gamma^{900}$ is an integer.
The original problem is equivalent to $|\beta^{900}+\gamma^{900}|<4^{-100}$
$\iff 2 \Re \beta^{900}<4^{-100}$
But we can prove $|\beta^9|<5^{-1}$ . See here
So we are done.

This post has been edited 2 times. Last edited by smy2012, Feb 25, 2017, 3:01 PM

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yayups

#7 h Feb 19, 2019, 7:26 AM • 4 Y

Y by mijail, Adventure10, Mango247, Mango247

Cute Problem:

Note that $\lambda$ is a root of $P(X)=(X-1)^3-X^2=X^3-4X^2+3X-1$ . It is easy to check that $P(X)$ has exactly one real root $\lambda$ , and it is between $3$ and $4$ . Let $\alpha,\bar{\alpha}$ be the other (complex) roots. We see that $|\alpha|^2\lambda=1$ , so $|\alpha|=1/\sqrt{\lambda}$ , so $1/2<|\alpha|<1/\sqrt{3}$ .

Let $x_n=\lambda^n+\alpha^n+\bar{\alpha}^n$ . Note by Newton sums or whatever that $x_n$ is always an integer. Now, let $M=x_{300}$ . We see that
$|M-\lambda^{300}|=|2\Re(\alpha^{300})|<2/3^{150}<4^{-100},$ as desired. $\blacksquare$

This post has been edited 1 time. Last edited by yayups, Feb 19, 2019, 8:09 AM

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Eyed

#8 h Jan 2, 2021, 6:00 PM

Y by

How to do this without a calculator help

Let $y = \lambda^{\frac{1}{3}}$ , we have the equation $y^{3} - y^{2} - 1 = 0$ . If $y$ was a real number, observe that $1.4^{3} - 1.4^{2} - 1 < 0$ but $1.5^{3} - 1.5^{2} - 1 > 0$ , so $1.4 < y < 1.5$ . I claim there are no other real solutions; for $y > 1.5$ then $y^{3} - y^{2} > 1$ , for $-1 < y < 1.4$ we have $y^{3} - y^{2} < 1$ , and for $y < -1$ we have $y^{3} - y^{2} < 1$ . Thus, the $1.4<y<1.5$ is our only real solution.

Let the two complex solutions be $z$ and $\overline{z}$ . Since $z\overline{z} = |z|^{2}$ , and the product of the roots is $1$ , we have $|z|^{2} < \frac{1}{1.4} = \frac{5}{7}$ . Consider the sum
$y^{900} + z^{900} + \overline{z}^{900}$ This is an integer since we can write it as the sum of symmmetric polynomials. Furthermore, observe that for any complex number $w$ , we have $w + \overline{w} < 2|w|$ , so
$z^{900} + \overline{z}^{900} < 2|z^{900}| < 2\cdot (\frac{5}{7})^{450} < 4^{-100}$ Thus, subtracting $z^{900} + \overline{z}^{900}$ from an integer results in $y^{900}$ , which has a fractional part greater than $1 - 4^{-100}$ , which proves the problem.

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1301 posts

#9 h May 27, 2022, 6:33 AM • 3 Y

Y by Mango247, Mango247, Mango247

Let $\lambda^{\frac 13}, r_1, r_2$ be the roots of $P(x)=x^3-x^2-1=0$ . Note $P(x)$ has a unique real root because $P(x)>0$ for all $x<1$ . We can see for $x\ge 1, P'(x)=3x^2-2x>0$ .

Therefore, $r_1, r_2$ are conjugates. We have $|r_1|^2=r_1r_2=\frac{1}{\lambda^{\frac 13}}$ .

We estimate $\lambda^{\frac 13}$ . Note $P(\sqrt{2})=2\sqrt{2}-2-1<0$ . Since $P$ is strictly increasing in $[1, \infty)$ it follows that $\lambda^{\frac 13}>\sqrt{2}$

Let $S_j=\lambda^{j/3}+r_1^j+r_2^j$ . Observe $S_j-S_{j-1}-S_{j-3}=0$ and $S_0=3, S_1=0, S_2=-1, S_3\in \mathbb{Z}$ .

I claim $M=S_{900}$ works. We have $|M-\lambda^{300}|=|r_1^{900}+r_2^{900}|\le 2(|r_1|^{2})^{450} \le 2 (\frac{1}{\sqrt{2}})^{450} \le 2^{1-225} \le 4^{-100}$ , as desired.

This post has been edited 1 time. Last edited by CANBANKAN, May 27, 2022, 2:47 PM

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#10 h May 28, 2022, 7:21 PM • 1 Y

Y by Mango247

At first place the $\lambda$ constant seems non-sense but to find a sense into it, u need to analyse the polynomial $P(x)=x^3-x^2-1$ , the problem gives us that $\lambda^{\frac{1}{3}}$ is a root of $P$ , now clearly $P(x)<0$ if $x<1$ so $\lambda>1$ , now $P'(x)=3x^2-2x$ which means that $P$ is strictly increasing if $x>\frac{2}{3}$ and strictly decreasing if $x<\frac{2}{3}$ ,hence if $P$ had all of its roots real then all of these roots will be greater than $1$ but $P$ is strictly increasing so we cant have $3$ different values of $P$ being the same. Which means that $P$ has $1$ real root and $2$ complex roots. So let $P(\omega)=0$ and $P(\overline{\omega})=0$ be the complex roots of $P$ and now by Newton Sums we have that $\lambda^{300}+\omega^{900}+\overline{\omega}^{900}$ is an positive integer, now it remains to show that this is the $M$ we are looking for, so the problem is equivalent to showing that $|2 \cdot \text{Re}(\omega^{900})|<4^{-100}$ , now let $\omega=a+bi$ then we need $2a^{900}<4^{-100}$ , but by vieta's formulas we have that $a^2<a^2+b^2=\frac{1}{\lambda^{\frac{1}{3}}}$ so we need to bound $\lambda$ , first lets try with the value $\sqrt{2}$ , $P(\sqrt{2})=2^{\frac{3}{2}}-3<0$ becuase $8<9$ so $\lambda>2^{\frac{3}{2}}$ since $P$ is strictly increasing for all $x>\frac{2}{3}$ which means that $a^2<\frac{1}{\lambda}<4^{-\frac{3}{4}}$ and now taking the $450$ -th power in both sides we get $a^{900}<4^{-\frac{1350}{4}}$ so $2a^{900}<4^{-\frac{1348}{4}}<4^{-100}$ and we are done

This post has been edited 2 times. Last edited by MathLuis, May 28, 2022, 7:26 PM

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#11 h Sep 8, 2023, 2:38 PM • 1 Y

Y by centslordm

Note that $\lambda^{1/3}$ is a real root of $x^3-x^2-1=0$ . By Descartes' rule of signs, this root is unique and positive, and we can check that it is greater than $1.44$ . Then by Vieta's this quadratic has two other complex roots $z,\overline{z}$ that are conjugates of each other, whose product is $1/\lambda^{1/3}$ , hence the magnitude of these roots (which is the same for both) is at most $1/\sqrt{1.44}=1/1.2$ . Then $1.2^9>5$ , so the magnitudes of $z^{900}$ and $\overline{z}^{900}$ are at most $5^{-100}$ , so their sum has magnitude less than $4^{-100}$ . On the other hand, $\lambda^{300}+z^{900}+\overline{z}^{900}$ is an integer, hence $\lambda^{300}$ is within $4^{-100}$ of some integer. $\blacksquare$

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1482 posts

#12 h Mar 13, 2025, 4:03 AM

Y by

Let $\lambda = x^3$ , $x > 0$ such that $x^3 = x^2 + 1$ and we want to show that $\left|M - x^{900}\right| < 4^{-100}$ Since $2\sqrt{2} < 3$ , we may note that the positive root $x$ of $x^3 - x^2 - 1$ has magnitude at least $\sqrt{2}$ . Since the other two roots then have negative sum and since this polynomial has no negative solutions, the other two roots $r_1, r_2$ are complex conjugates with magnitudes $R^{-1/2}$ .
Since $x^{900} + r_1^{900} + r_2^{900}$ is an integer by Newton's sums, it suffices to show that and thus if we show $R^{-900/2} < 2^{-201} \iff 2^{402} < R^{900}$ we will be done. Since $R > \sqrt{2}$ , it follows that $R^{900} > 2^{450}$ and we are done.

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