\gcd(3^{15} - 1, 3^9 - 1) = 3^{\gcd(15,9)} - 1.

February 22, 2026

$\gcd(3^{15} - 1, 3^9 - 1) = 3^{\gcd(15,9)} - 1.$

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Question: What is the greatest common divisor of $3^{15} - 1$ and $3^9 - 1$?
Solution: Use the property:
\gcd(a^m - 1, a^n - 1) = a^{\gcd(m,n)} - 1, \quad ext{for } a > 1.
Now compute $\gcd(15,9) = 3$.
\gcd(3^{15} - 1, 3^9 - 1) = 3^3 - 1 = 27 - 1 = 26.
Therefore, the greatest common divisor is $oxed{26}$.**Question:** An archaeologist discovers an ancient Egyptian artifact with markings that represent two numbers \(x\) and \(y\), which are believed to encode a relationship through the equation: \[rac{x+y}{x-y} + rac{x-y}{x+y} = 3.\] Find all possible pairs \((x, y)\) of real numbers satisfying this equation.

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