query: Can someone explain to me why the greatest common factor of (a,b) is equal to gcf(a mod b, a)
positive_passages.docid: 190
positive_passages.text: Why $\gcd(qb+r,b)=\gcd(b,r)$?
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: 84417
negative_passages.text: Prove that if $\gcd(a,b)=1$ then $\gcd(a^m, b)=1$
negative_passages.docid: 117538
negative_passages.text: What is the interpretation of $a \equiv b$ mod $H$ in group theory?
negative_passages.docid: 60842
negative_passages.text: if $a^2+b^2=c^2$ then $gcd(a,b)=gcd(a,c)=gcd(b,c)$.
negative_passages.docid: 1991
negative_passages.text: If $\gcd(a,b)=1$ and $a$ and $b$ divide $c$, then so does $ab$
negative_passages.docid: 154959
negative_passages.text: Proof of associativity of GCD?
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: 69279
negative_passages.text: If $\gcd(a,b)=1$ then, $\gcd(a^2,b^2)=1$
negative_passages.docid: 19656
negative_passages.text: Proving that $ \gcd(a,b) = as + bt $, i.e., $ \gcd $ is a linear combination.
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: 96535
negative_passages.text: Show that if $\gcd(a,3)=1$ then $a^7 \equiv a\pmod{63}$. Why is this assumption necessary?
negative_passages.docid: 102133
negative_passages.text: Prove: $\gcd(a,b) = \gcd(a, b + at)$.
negative_passages.docid: 78678
negative_passages.text: If $\gcd(m,n)=1$, and $\gcd(a,m)=\gcd(b,n)=1$, then $\gcd(ab, mn)=1$
negative_passages.docid: 20823
negative_passages.text: If $\gcd(a,b) = 1$ and $a,b\mid x$ then $ab\mid x$.
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: 92518
negative_passages.text: mod Distributive Law, factoring $\!\!\bmod\!\!:$ $\ ab\bmod ac = a(b\bmod c)$