query: Prove $\gcd(k, l) = d \Rightarrow \gcd(2^k - 1, 2^l - 1) = 2^d - 1$
positive_passages.docid: 6012
positive_passages.text: Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$
negative_passages.docid: 21788
negative_passages.text: Proving that if $ad-bc = \pm 1$, then $\gcd(x,y) = \gcd(ax +by, cx + dy)$
negative_passages.docid: 31465
negative_passages.text: Prove that if $a \mid b$ then $\gcd(a,b) = |a|$
negative_passages.docid: 59479
negative_passages.text: For a positive integer $n\geq 2$ with divisors $1=d_1<d_2<\cdots<d_k=n$, prove that $d_1d_2+d_2d_3+\cdots+d_{k-1}d_k<n^2$
negative_passages.docid: 102133
negative_passages.text: Prove: $\gcd(a,b) = \gcd(a, b + at)$.
negative_passages.docid: 63547
negative_passages.text: Euclid's Lemma and $\gcd(a,b)=1\Rightarrow \gcd(a^n,b^k)= 1$
negative_passages.docid: 20574
negative_passages.text: Understanding the proof for $\gcd (F_n, F_{n+k}) = 1$ where $F_n$ is a Fermat number
negative_passages.docid: 83307
negative_passages.text: If $n\mid a^2-1$, $n$ an odd integer, show $n = \gcd(a-1, n)\gcd(a+1, n)$
negative_passages.docid: 69279
negative_passages.text: If $\gcd(a,b)=1$ then, $\gcd(a^2,b^2)=1$
negative_passages.docid: 82938
negative_passages.text: $\gcd(a,b)=1, x^a = y^b\Rightarrow x = n^b$, $ y = n^a$ for an integer $n$.
negative_passages.docid: 34330
negative_passages.text: How to prove $\gcd(a^2,b^2) = (\gcd(a,b))^2$?
negative_passages.docid: 126358
negative_passages.text: Prove for $\gcd$ of two integers $a,b$, that $(a^n,b^n) = (a,b)^n$
negative_passages.docid: 38404
negative_passages.text: Prove $\gcd(a+b, a-b) = 1$ or $2\,$ if $\,\gcd(a,b) = 1$
negative_passages.docid: 37808
negative_passages.text: Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$
negative_passages.docid: 53777
negative_passages.text: Why is $\gcd(x^4+1,x^2-1) = 1$ but I get $2$? [unit normalization of gcds]
negative_passages.docid: 72424
negative_passages.text: Suppose 1 ≤ k ≤ n − 1 and gcd(k, n) = 1. Prove that gcd(n − k, n) = 1
negative_passages.docid: 2289
negative_passages.text: Prove that if $\gcd( a, b ) = 1$ then $\gcd( ac, b ) = \gcd( c, b ) $