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Difference between revisions of "2020 CIME II Problems/Problem 9"

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==Problem 9==
 
==Problem 9==
 
Let <math>f(x)=x^2-2</math>. There are <math>N</math> real numbers <math>x</math> such that <cmath>\underbrace{f(f(\ldots f}_{2019\text{ times}}(x)\ldots))=\underbrace{f(f(\ldots f}_{2020\text{ times}}(x)\ldots)).</cmath>Find the remainder when <math>N</math> is divided by <math>1000</math>.
 
Let <math>f(x)=x^2-2</math>. There are <math>N</math> real numbers <math>x</math> such that <cmath>\underbrace{f(f(\ldots f}_{2019\text{ times}}(x)\ldots))=\underbrace{f(f(\ldots f}_{2020\text{ times}}(x)\ldots)).</cmath>Find the remainder when <math>N</math> is divided by <math>1000</math>.
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==Solution==
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==Solution 1==
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Let <math>f^n(x)</math> denote the composition of <math>f</math> with itself <math>n</math> times. We require that <math>f^{2019}(x)=f^{2020}(x)</math>. Since
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<math>f^{2020}(x)=f(f^{2019}(x))=(f^{2019}(x))^2-2</math>, we have <math>f^{2019}(x)=(f^{2019}(x))^2-2</math>. Hence
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<math>(f^{2019}(x)+1)(f^{2019}(x)-2)=0</math> so that <math>f^{2019}(x)\in\{-1,2\}</math>.
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Now if <math>f^{n+1}(x)=k</math> for some <math>k</math>, then <math>k=f^{n+1}(x)=f(f^{n}(x))=(f^{n}(x))^2-2</math> so that
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<math>f^{n}(x)=\pm\sqrt{k+2}</math>. Hence we find  <math>f^{2018}(x)\in\{\pm1,\pm2\}</math>, and similarly
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<math>f^{2017}(x)\in\{\pm\sqrt{3},\pm1,\pm2,0\}</math>. Note that at each step since our values for <math>k</math> are <math>\geq -2</math> we have
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<math>\sqrt{k+2}</math> is real; and <math>\pm\sqrt{k+2}</math> are distinct, and also <math>\neq k</math>, except when <math>k=-2</math>. Note that every <math>x</math> that satisfies <math>f^{n+1}(x)=k</math> also satisfies <math>f^{n}(x)=\pm\sqrt{k+2}</math>. Since <math>-2</math> is a value of <math>k</math> starting with <math>f^{2018}</math>, we have for <math>n\leq 2018</math>, <math>f^{n}(x)</math> must equal any one of <math>3\cdot 2^{2018-n}+1</math> many values. It follows that there are <math>3\cdot 2^{2018}+1</math> many real solutions.
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By Eulers Theorem <math>2^{\phi(125)}=2^{100}\equiv1\mod 125</math> so <math>2^{2000}=(2^{100})^{20}\equiv 1^{20}=1\mod 125</math> and hence <math>2^{2003}\equiv 8\mod 1000</math>. Since <math>2^{15}=2^{10}\cdot 2^5\equiv 24\cdot 32=768\mod 1000</math> we have <math>2^{2018}\equiv 8\cdot 768\equiv 144\mod 1000</math>. Hence <math>3\cdot 2^{2018}+1\equiv 3\cdot 144+1=433\mod 1000</math>. Thus the answer is 433 as desired.
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==Solution 2==
 
We can start by finding the number of solution for smaller repeptitions of <math>f</math>. Notice that we can solve <math>f(f(x))=f(x)</math> by applying the functional inverse <math>f^{-1}</math> to both sides as you would to solve any equation: <math>f^{-1}(f(f(x)))=f^{-1}(f(x))\Longrightarrow f(x)=|x|</math> (We put the absolute value bars because we know that taking the inverse of <math>f</math> of both sides involves taking the square root of both sides, and <math>\sqrt{t^2}=|t|</math>). From here, it is easy to see that this equation has <math>4</math> solutions at <math>\pm1</math> and <math>\pm2</math>. We can also try for <math>f(f(f(x)))=f(f(x))</math> (we will solve more methodically here): <cmath>((x^2-2)^2-2)^2-2=(x^2-2)^2-2</cmath> <cmath>(x^2-2)^2-2)^2 = (x^2-2)^2</cmath> <cmath>(x^2-2)^2-2=|x^2-2|</cmath> <cmath>(x^2-2)^2-2=\pm(x^2-2)</cmath> <cmath>(x^2-2)^2=x^2 \textup{  OR  } (x^2-2)^2=-x^2+4</cmath>The first equation yeilds <math>4</math> results, and the second equation yields <math>2</math> results for a total of <math>6</math> results. It appears that <math>\underbrace{f(f\cdots f}_{n\textup{ times}}(x))=\underbrace{f(f\cdots f}_{n-1\textup{ times}}(x))</math> bas <math>2n</math> real solutions, giving a total of <math>4040</math> apparent solutions for the original equation. This makes logical sense considering that <math>f</math> is an even polynomial with 2 roots. For a more formal proof, we consider <math>F_n(x)=\underbrace{f(f\cdots f}_{n\textup{ times}}(x))</math>. We are asked to find the number of solutions of the equation in the form <math>F_n(x)=F_{n-1}(x)</math>. Following from how we solved the first simple case, <math>f^{-1}(F_n(x))=f^{-1}(F_{n-1}(x)) = F_{n-1}(x)=|F_{n-2}(x)|</math>. Note that the absolute value branches off in rwo directions: <math>F_{n-1}(x)=\pm F_{n-2}(x)</math>. This would give a total of <math>2\cdot2n=4n</math> real and complex solutions (we multiply by 2 because <math>f</math> is a quadratic, which has 2 total roots). The complex roots come from the negative branches, so there are <math>2\cdot n=2n</math> complex solutions. Therefore, there are a total of <math>2n</math> real roots, which again gives <math>4040</math> roots for the original question. The question asks for <math>\boxed{040}\equiv4040(\mod1000)<cmath></cmath></math>~bhargavakanakapura
 
We can start by finding the number of solution for smaller repeptitions of <math>f</math>. Notice that we can solve <math>f(f(x))=f(x)</math> by applying the functional inverse <math>f^{-1}</math> to both sides as you would to solve any equation: <math>f^{-1}(f(f(x)))=f^{-1}(f(x))\Longrightarrow f(x)=|x|</math> (We put the absolute value bars because we know that taking the inverse of <math>f</math> of both sides involves taking the square root of both sides, and <math>\sqrt{t^2}=|t|</math>). From here, it is easy to see that this equation has <math>4</math> solutions at <math>\pm1</math> and <math>\pm2</math>. We can also try for <math>f(f(f(x)))=f(f(x))</math> (we will solve more methodically here): <cmath>((x^2-2)^2-2)^2-2=(x^2-2)^2-2</cmath> <cmath>(x^2-2)^2-2)^2 = (x^2-2)^2</cmath> <cmath>(x^2-2)^2-2=|x^2-2|</cmath> <cmath>(x^2-2)^2-2=\pm(x^2-2)</cmath> <cmath>(x^2-2)^2=x^2 \textup{  OR  } (x^2-2)^2=-x^2+4</cmath>The first equation yeilds <math>4</math> results, and the second equation yields <math>2</math> results for a total of <math>6</math> results. It appears that <math>\underbrace{f(f\cdots f}_{n\textup{ times}}(x))=\underbrace{f(f\cdots f}_{n-1\textup{ times}}(x))</math> bas <math>2n</math> real solutions, giving a total of <math>4040</math> apparent solutions for the original equation. This makes logical sense considering that <math>f</math> is an even polynomial with 2 roots. For a more formal proof, we consider <math>F_n(x)=\underbrace{f(f\cdots f}_{n\textup{ times}}(x))</math>. We are asked to find the number of solutions of the equation in the form <math>F_n(x)=F_{n-1}(x)</math>. Following from how we solved the first simple case, <math>f^{-1}(F_n(x))=f^{-1}(F_{n-1}(x)) = F_{n-1}(x)=|F_{n-2}(x)|</math>. Note that the absolute value branches off in rwo directions: <math>F_{n-1}(x)=\pm F_{n-2}(x)</math>. This would give a total of <math>2\cdot2n=4n</math> real and complex solutions (we multiply by 2 because <math>f</math> is a quadratic, which has 2 total roots). The complex roots come from the negative branches, so there are <math>2\cdot n=2n</math> complex solutions. Therefore, there are a total of <math>2n</math> real roots, which again gives <math>4040</math> roots for the original question. The question asks for <math>\boxed{040}\equiv4040(\mod1000)<cmath></cmath></math>~bhargavakanakapura

Revision as of 08:32, 2 January 2022

Problem 9

Let $f(x)=x^2-2$. There are $N$ real numbers $x$ such that \[\underbrace{f(f(\ldots f}_{2019\text{ times}}(x)\ldots))=\underbrace{f(f(\ldots f}_{2020\text{ times}}(x)\ldots)).\]Find the remainder when $N$ is divided by $1000$.

Solution 1

Let $f^n(x)$ denote the composition of $f$ with itself $n$ times. We require that $f^{2019}(x)=f^{2020}(x)$. Since

$f^{2020}(x)=f(f^{2019}(x))=(f^{2019}(x))^2-2$, we have $f^{2019}(x)=(f^{2019}(x))^2-2$. Hence

$(f^{2019}(x)+1)(f^{2019}(x)-2)=0$ so that $f^{2019}(x)\in\{-1,2\}$.


Now if $f^{n+1}(x)=k$ for some $k$, then $k=f^{n+1}(x)=f(f^{n}(x))=(f^{n}(x))^2-2$ so that

$f^{n}(x)=\pm\sqrt{k+2}$. Hence we find $f^{2018}(x)\in\{\pm1,\pm2\}$, and similarly

$f^{2017}(x)\in\{\pm\sqrt{3},\pm1,\pm2,0\}$. Note that at each step since our values for $k$ are $\geq -2$ we have

$\sqrt{k+2}$ is real; and $\pm\sqrt{k+2}$ are distinct, and also $\neq k$, except when $k=-2$. Note that every $x$ that satisfies $f^{n+1}(x)=k$ also satisfies $f^{n}(x)=\pm\sqrt{k+2}$. Since $-2$ is a value of $k$ starting with $f^{2018}$, we have for $n\leq 2018$, $f^{n}(x)$ must equal any one of $3\cdot 2^{2018-n}+1$ many values. It follows that there are $3\cdot 2^{2018}+1$ many real solutions.

By Eulers Theorem $2^{\phi(125)}=2^{100}\equiv1\mod 125$ so $2^{2000}=(2^{100})^{20}\equiv 1^{20}=1\mod 125$ and hence $2^{2003}\equiv 8\mod 1000$. Since $2^{15}=2^{10}\cdot 2^5\equiv 24\cdot 32=768\mod 1000$ we have $2^{2018}\equiv 8\cdot 768\equiv 144\mod 1000$. Hence $3\cdot 2^{2018}+1\equiv 3\cdot 144+1=433\mod 1000$. Thus the answer is 433 as desired.

Solution 2

We can start by finding the number of solution for smaller repeptitions of $f$. Notice that we can solve $f(f(x))=f(x)$ by applying the functional inverse $f^{-1}$ to both sides as you would to solve any equation: $f^{-1}(f(f(x)))=f^{-1}(f(x))\Longrightarrow f(x)=|x|$ (We put the absolute value bars because we know that taking the inverse of $f$ of both sides involves taking the square root of both sides, and $\sqrt{t^2}=|t|$). From here, it is easy to see that this equation has $4$ solutions at $\pm1$ and $\pm2$. We can also try for $f(f(f(x)))=f(f(x))$ (we will solve more methodically here): \[((x^2-2)^2-2)^2-2=(x^2-2)^2-2\] \[(x^2-2)^2-2)^2 = (x^2-2)^2\] \[(x^2-2)^2-2=|x^2-2|\] \[(x^2-2)^2-2=\pm(x^2-2)\] \[(x^2-2)^2=x^2 \textup{ OR } (x^2-2)^2=-x^2+4\]The first equation yeilds $4$ results, and the second equation yields $2$ results for a total of $6$ results. It appears that $\underbrace{f(f\cdots f}_{n\textup{ times}}(x))=\underbrace{f(f\cdots f}_{n-1\textup{ times}}(x))$ bas $2n$ real solutions, giving a total of $4040$ apparent solutions for the original equation. This makes logical sense considering that $f$ is an even polynomial with 2 roots. For a more formal proof, we consider $F_n(x)=\underbrace{f(f\cdots f}_{n\textup{ times}}(x))$. We are asked to find the number of solutions of the equation in the form $F_n(x)=F_{n-1}(x)$. Following from how we solved the first simple case, $f^{-1}(F_n(x))=f^{-1}(F_{n-1}(x)) = F_{n-1}(x)=|F_{n-2}(x)|$. Note that the absolute value branches off in rwo directions: $F_{n-1}(x)=\pm F_{n-2}(x)$. This would give a total of $2\cdot2n=4n$ real and complex solutions (we multiply by 2 because $f$ is a quadratic, which has 2 total roots). The complex roots come from the negative branches, so there are $2\cdot n=2n$ complex solutions. Therefore, there are a total of $2n$ real roots, which again gives $4040$ roots for the original question. The question asks for $\boxed{040}\equiv4040(\mod1000)<cmath></cmath>$~bhargavakanakapura

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