2021 AIME II Problems/Problem 9

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 (Generalized and Comprehensive)

Claim

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

~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 $p>q>0,$ we have $\gcd(p,q)=\gcd(p-q,q).$ We apply this result repeatedly to reduce the larger number: $\gcd\left(u^a-1,u^b-1\right)=\gcd\left(u^b-1,u^a-1-u^{a-b}\left(u^b-1\right)\right)=\gcd\left(u^b-1,u^{a-b}-1\right).$

~MRENTHUSIASM

Proof 2

Solution in progress. A million thanks for not editing.

~MRENTHUSIASM

Solution

Solution in progress. A million thanks for not editing.

~MRENTHUSIASM

2021 AIME II (Problems • Answer Key • Resources)
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2021 AIME II Problems/Problem 9

Contents

Problem

Solution 1

Solution 2 (Generalized and Comprehensive)

Claim

Proof 1 (Euclidean Algorithm)

Proof 2

Solution

See also