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Difference between revisions of "2021 AIME II Problems/Problem 9"

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Let <math>d=\gcd\left(u^a-1,u^b-1\right).</math> It follows that <math>u^a\equiv1\pmod{d}</math> and <math>u^b\equiv1\pmod{d}.</math>
 
Let <math>d=\gcd\left(u^a-1,u^b-1\right).</math> It follows that <math>u^a\equiv1\pmod{d}</math> and <math>u^b\equiv1\pmod{d}.</math>
  
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By Bézout's Identity, there exist integers <math>x</math> and <math>y</math> for which <math>ax+by=\gcd(a,b),</math> thus <cmath>u^{\gcd(a,b)}=u^{ax+by}=\left(u^a\right)^x\left(u^b\right)^y\equiv1\pmod{d},</cmath>
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By Bézout's Identity, there exist integers <math>x</math> and <math>y</math> for which <math>ax+by=\gcd(a,b),</math> so <cmath>u^{\gcd(a,b)}=u^{ax+by}=\left(u^a\right)^x\left(u^b\right)^y\equiv1\pmod{d},</cmath>
 
from which <math>u^{\gcd(a,b)}-1\equiv0\pmod{d}.</math> We know that <math>u^{\gcd(a,b)}-1\geq d.</math>
 
from which <math>u^{\gcd(a,b)}-1\equiv0\pmod{d}.</math> We know that <math>u^{\gcd(a,b)}-1\geq d.</math>
  

Revision as of 22:45, 2 April 2021

Problem

Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$.

Solution 1

We make use of the (olympiad number theory) lemma that $\gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1$.

Noting $\gcd(2^m+1,2^m-1)=\gcd(2^m+1,2)=1$, we have (by difference of squares)\[\gcd(2^m+1,2^n-1) \neq 1 \iff \gcd(2^{2m}-1,2^n-1) \neq \gcd(2^m-1,2^n-1)\]\[\iff 2^{\gcd(2m,n)}-1 \neq 2^{\gcd(m,n)}-1 \iff \gcd(2m,n) \neq \gcd(m,n) \iff \nu_2(m)<\nu_2(n).\] It is now easy to calculate the answer (with casework on $\nu_2(m)$) as $15 \cdot 15+8 \cdot 7+4 \cdot 3+2 \cdot 1=\boxed{295}$.

~Lcz

Solution 2 (Comprehensive and Generalized)

Claim (Solution 1's Lemma)

If $u,a,$ and $b$ are positive integers with $u\geq2,$ then $\gcd\left(u^a-1,u^b-1\right)=u^{\gcd(a,b)}-1.$

There are two proofs to this claim, as shown below.

~MRENTHUSIASM

Proof 1 (Euclidean Algorithm)

If $a=b,$ then $\gcd(a,b)=a=b,$ from which the claim is clearly true.

Otherwise, let $a>b$ without the loss of generality. Note that for all integers $p>q>0,$ the Euclidean Algorithm states that \[\gcd(p,q)=\gcd(p-q,q)=\cdots=\gcd(q,p\text{ mod }q).\] We apply this result repeatedly to reduce the larger number: \[\gcd\left(u^a-1,u^b-1\right)=\gcd\left(u^b-1,u^a-1-u^{a-b}\left(u^b-1\right)\right)=\gcd\left(u^b-1,u^{a-b}-1\right).\] Continuing, we have \begin{align*} \gcd\left(u^a-1,u^b-1\right)&=\cdots \\ &=\gcd\left(u^b-1,u^{a-b}-1\right) \\ &=\cdots \\ &=\gcd\left(u^{\gcd(a,b)}-1,u^{\gcd(a,b)}-1\right) \\ &=u^{\gcd(a,b)}-1, \end{align*} from which the proof is complete.

~MRENTHUSIASM

Proof 2 (Bézout's Identity)

Let $d=\gcd\left(u^a-1,u^b-1\right).$ It follows that $u^a\equiv1\pmod{d}$ and $u^b\equiv1\pmod{d}.$

By Bézout's Identity, there exist integers $x$ and $y$ for which $ax+by=\gcd(a,b),$ so \[u^{\gcd(a,b)}=u^{ax+by}=\left(u^a\right)^x\left(u^b\right)^y\equiv1\pmod{d},\] from which $u^{\gcd(a,b)}-1\equiv0\pmod{d}.$ We know that $u^{\gcd(a,b)}-1\geq d.$

Next, we notice that \begin{align*} u^a-1&=\left(u^{\gcd(a,b)}-1\right)\left(u^{a-\gcd{(a,b)}}+u^{a-2\gcd{(a,b)}}+u^{a-3\gcd{(a,b)}}+\cdots+1\right), \\ u^b-1&=\left(u^{\gcd(a,b)}-1\right)\left(u^{b-\gcd{(a,b)}}+u^{b-2\gcd{(a,b)}}+u^{b-3\gcd{(a,b)}}+\cdots+1\right). \end{align*} Since $u^{\gcd(a,b)}-1$ is a common divisor of $u^a-1$ and $u^b-1,$ we conclude that $u^{\gcd(a,b)}-1=d,$ from which the proof is complete.

~MRENTHUSIASM

Solution (Detailed Explanation of Solution 1)

By the Euclidean Algorithm, we have \[\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2^m-1,2\right)=1. \hspace{41.125mm} (*)\] We are given that $\gcd\left(2^m+1,2^n-1\right)>1.$ Multiplying both sides by $\gcd\left(2^m-1,2^n-1\right)$ gives \begin{align*} \gcd\left(2^m+1,2^n-1\right)\cdot\gcd\left(2^m-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \hspace{5mm} (**) \\ \underbrace{\gcd\left(\left(2^m+1\right)\left(2^m-1\right),2^n-1\right)}_{\text{by }(*)\text{ and }\textbf{Remark (GCD Property)}}&>\gcd\left(2^m-1,2^n-1\right) \\ \gcd\left(2^{2m}-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ \underbrace{2^{\gcd(2m,n)}-1}_{\text{by }\textbf{Claim}}&>\underbrace{2^{\gcd(m,n)}-1}_{\text{by }\textbf{Claim}} \\ \gcd(2m,n)&>\gcd(m,n), \end{align*} which implies that $n$ must have more factors of $2$ than $m$ does.

We construct the following table for the first $30$ positive integers: \[\begin{array}{c|c|c} && \\ [-2.5ex] \textbf{\# of Factors of }2 & \textbf{Numbers} & \textbf{Count} \\ \hline && \\ [-2.25ex] 0 & 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 & 15 \\ && \\ [-2.25ex] 1 & 2,6,10,14,18,22,26,30 & 8 \\ && \\ [-2.25ex] 2 & 4,12,20,28 & 4 \\ && \\ [-2.25ex] 3 & 8,24 & 2 \\ && \\ [-2.25ex] 4 & 16 & 1 \\ \end{array}\] To count the ordered pairs $(m,n),$ we perform casework on the number of factors of $2$ that $m$ has:

  1. If $m$ has $0$ factors of $2,$ then $m$ has $15$ options and $n$ has $8+4+2+1=15$ options. So, this case has $15\cdot15=225$ ordered pairs.

  2. If $m$ has $1$ factor of $2,$ then $m$ has $8$ options and $n$ has $4+2+1=7$ options. So, this case has $8\cdot7=56$ ordered pairs.

  3. If $m$ has $2$ factors of $2,$ then $m$ has $4$ options and $n$ has $2+1=3$ options. So, this case has $4\cdot3=12$ ordered pairs.

  4. If $m$ has $3$ factors of $2,$ then $m$ has $2$ options and $n$ has $1$ option. So, this case has $2\cdot1=2$ ordered pairs.

Together, the answer is $225+56+12+2=\boxed{295}.$

~MRENTHUSIASM

Remark (GCD Property)

In $(**)$ of the solution, we use the following property of the greatest common divisor:

For positive integers $r,s,$ and $t,$ if $\gcd(r,s)=1,$ then $\gcd(r,t)\cdot\gcd(s,t)=\gcd(rs,t).$

As $r$ and $s$ are relatively prime (have no prime divisors in common), this property is intuitive.

~MRENTHUSIASM

See Also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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