query: Prove that $\gcd (a, b) = \gcd (a, b + xa)$ for any $x \in \mathbb{Z}$.
positive_passages.docid: 190
positive_passages.text: Why $\gcd(qb+r,b)=\gcd(b,r)$?
negative_passages.docid: 105832
negative_passages.text: prove that if $\gcd(a,b)=1$ then $\gcd(a^2, b^2)=1$
negative_passages.docid: 31331
negative_passages.text: $|a|=m,\,\gcd(m,n)=1 \implies a$ is an $n$'th power
negative_passages.docid: 7091
negative_passages.text: If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
negative_passages.docid: 84417
negative_passages.text: Prove that if $\gcd(a,b)=1$ then $\gcd(a^m, b)=1$
negative_passages.docid: 72747
negative_passages.text: Proving gcd($a,b$)lcm($a,b$) = $|ab|$
negative_passages.docid: 80148
negative_passages.text: Prove $\gcd(a+b,a^2+b^2)$ is $1$ or $2$ if $\gcd(a,b) = 1$
negative_passages.docid: 23066
negative_passages.text: If $\gcd(a, b) = 1$, then $\gcd(ab, c) = \gcd(a, c) \cdot\gcd(b, c)$
negative_passages.docid: 55673
negative_passages.text: Alternate proof that $\exists\ x \in \Bbb{Z}$ such that $ \gcd (a+bx,c) = 1$?
negative_passages.docid: 7090
negative_passages.text: Show that $\gcd(a,bc)=1$ if and only if $\gcd(a,b)=1$ and $\gcd(a,c)=1$
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: 19656
negative_passages.text: Proving that $ \gcd(a,b) = as + bt $, i.e., $ \gcd $ is a linear combination.
negative_passages.docid: 69967
negative_passages.text: How to prove $\gcd(a^m-b^m,a^n-b^n) = a^{\gcd(m,n)} - b^{\gcd(m,n)} $?
negative_passages.docid: 15168
negative_passages.text: If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$. and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $
negative_passages.docid: 149993
negative_passages.text: Use the Euclidean algorithm to prove that gcd(na, nb) = n gcd(a, b).
negative_passages.docid: 32344
negative_passages.text: Let $a,k,m$ be integers. Prove that $\gcd(ka,km) = k\gcd(a,m)$.
negative_passages.docid: 76507
negative_passages.text: $\gcd(b^x - 1, b^y - 1, b^ z- 1,\dots) = b^{\gcd(x, y, z,\dots)} -1$