Let $m$ and $n$ be relatively prime positive integers. Determine all possible values of

$$
\operatorname{gcd}\left(2^{m}-2^{n}, 2^{m^{2}+m n+n^{2}}-1\right) .
$$

Answer: 1 and 7.
Without restriction of generality we may assume that \( m \geqslant n \). It is well known that

\[
\operatorname{gcd}\left(2^{p}-1,2^{q}-1\right)=2^{\operatorname{gcd}(p, q)}-1,
\]

so

\[
\begin{aligned}
\operatorname{gcd}\left(2^{m}-2^{n}, 2^{m^{2}+m n+n^{2}}-1\right) & =\operatorname{gcd}\left(2^{m-n}-1,2^{m^{2}+m n+n^{2}}-1\right) \\
& =2^{\operatorname{gcd}\left(m-n, m^{2}+m n+n^{2}\right)}-1 .
\end{aligned}
\]

Let \( d \geqslant 1 \) be a divisor of \( m-n \). Clearly, \(\operatorname{gcd}(m, d)=1\), since \( m \) and \( n \) are relatively prime. Assume that \( d \) also divides \( m^{2}+m n+n^{2} \). Then \( d \) divides \( 3 m^{2} \). In view of \(\operatorname{gcd}(m, d)=1\) the only choices for \( d \) are \( d=1 \) and \( d=3 \). Hence \(\operatorname{gcd}\left(m-n, m^{2}+m n+n^{2}\right)\) is either 1 or 3. Consequently, \(\operatorname{gcd}\left(2^{m}-2^{n}, 2^{m^{2}+m n+n^{2}}-1\right)\) may only assume the values \( 2^{1}-1=1 \) and \( 2^{3}-1=7 \).

Both values are attainable, since \( m=2, n=1 \) gives

\[
\operatorname{gcd}\left(2^{2}-2^{1}, 2^{2^{2}+2 \cdot 1+1^{2}}-1\right)=\operatorname{gcd}\left(2,2^{7}-1\right)=1
\]

whereas \( m=1, n=1 \) gives

\[
\operatorname{gcd}\left(2^{1}-2^{1}, 2^{1^{2}+1 \cdot 1+1^{2}}-1\right)=\operatorname{gcd}\left(0,2^{3}-1\right)=7
\]