query: Proving $\gcd( m,n)$=1
positive_passages.docid: 18714
positive_passages.text: $\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1$, or $p$
negative_passages.docid: 23590
negative_passages.text: Show that $\gcd\left(\frac{a^n-b^n}{a-b},a-b\right)=\gcd(n d^{n-1},a-b)$
negative_passages.docid: 63965
negative_passages.text: How can I compute the sum of $ {m\over\gcd(m,n)}$?
negative_passages.docid: 31331
negative_passages.text: $|a|=m,\,\gcd(m,n)=1 \implies a$ is an $n$'th power
negative_passages.docid: 42780
negative_passages.text: Show that if $a^{n-1} \equiv 1 \pmod{n}$ for all $a$ such that $\gcd(n,a) = 1$, then $a^{n} \equiv a \pmod{n}$ for all $a$.
negative_passages.docid: 105832
negative_passages.text: prove that if $\gcd(a,b)=1$ then $\gcd(a^2, b^2)=1$
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: 10348
negative_passages.text: Prove that if the $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1$
negative_passages.docid: 1197
negative_passages.text: Finding $\gcd(2n-1,2n+1)$ . . . I think it's 1 but is my logic right?
negative_passages.docid: 157508
negative_passages.text: prove $lcm(a,gcd(b,c))=gcd(lcm(a,b),lcm(a,c))$
negative_passages.docid: 138058
negative_passages.text: Prove that $\mathbb Z_{m}\times\mathbb Z_{n} \cong \mathbb Z_{mn}$ implies $\gcd(m,n)=1$.
negative_passages.docid: 136064
negative_passages.text: How can I prove that $\gcd(a,b)=1\implies \gcd(a^2,b^2)=1$ without using prime decomposition?
negative_passages.docid: 74924
negative_passages.text: identity element of $(D_n ,\gcd)$
negative_passages.docid: 73972
negative_passages.text: How to prove that $\gcd(2n^3-14n+2,n+3) = \gcd(n+3,10)$?
negative_passages.docid: 4141
negative_passages.text: Prove: If $a\mid m$ and $b\mid m$ and $\gcd(a,b)=1$ then $ab\mid m$
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: 7151
negative_passages.text: Let $a,m,n \in \mathbf{N}$. Show that if $\gcd(m,n)=1$, then $\gcd(a,mn)=\gcd(a,m)\cdot\gcd(a,n)$.