query: prove that if $u \equiv w \bmod \Phi(m)$ and $\operatorname{gcd}(a, m)=1$ then $a^{u} \equiv a^{w} \bmod m$
positive_passages.docid: 4972
positive_passages.text: Mod of numbers with large exponents [modular order reduction]
negative_passages.docid: 110313
negative_passages.text: Prove that $\gcd(f(x),g(x)) = 1$.
negative_passages.docid: 139659
negative_passages.text: Find all $n$ with $\!\bmod n\!:\, a\,$ invertible $\Rightarrow a^{-1} \equiv a,\,$ i.e. $a^2\equiv 1$
negative_passages.docid: 14693
negative_passages.text: $d = \gcd(c,m),\ ac≡bc\pmod{\! m}\Rightarrow a≡b\pmod {\!m/d}\ $ [Congruence Cancellation & Division Rule]
negative_passages.docid: 62382
negative_passages.text: I suppose $(a,m)=(a-1,m)=1$, show that $1+a+a^2+\ldots+a^{\phi(m)-1}\equiv0\pmod m$
negative_passages.docid: 6410
negative_passages.text: Prove that $(ma, mb) = |m|(a, b)\ $ [GCD & LCM Distributive Law]
negative_passages.docid: 50396
negative_passages.text: How can I show that $ab \sim \gcd (a,b) {\operatorname{lcm} (a,b)}$ for any $a,b \in R \setminus \{0\}$?
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: 154959
negative_passages.text: Proof of associativity of GCD?
negative_passages.docid: 108450
negative_passages.text: Prove that $\gcd(a, b) = \gcd(a, b + ma)$?
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: 105832
negative_passages.text: prove that if $\gcd(a,b)=1$ then $\gcd(a^2, b^2)=1$
negative_passages.docid: 110752
negative_passages.text: $g^\frac{p-1}{2} \equiv -1 \ (mod \ p)$
negative_passages.docid: 7091
negative_passages.text: If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
negative_passages.docid: 108237
negative_passages.text: Show $a^p \equiv b^p \mod p^2$
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: 29204
negative_passages.text: $a\equiv\bar a\!\pmod{\!kn}\Rightarrow a\equiv\bar a\!\pmod{\! n};\ $ $(a\bmod kn)\bmod n=a\bmod n.\ $ Congruences persist mod factors of the modulus