How to show that $$\gcd \Biggl( {m \choose k} , {m+1 \choose k } , {m+2 \choose k}, \cdots, {m+k \choose k} \Biggr)=1$$
where $m,k \in \mathbb{N}$ and $m \geq k $.
Asked
Active
Viewed 282 times
2
-
12$\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$ is useful here. – J. M. ain't a mathematician Oct 20 '10 at 17:45
-
@J. M. very clever – Ross Millikan Oct 20 '10 at 18:06
1 Answers
1
We apply $\ \gcd(\cdots, \color{#0a0}a,\color{#c00}b) = \gcd(\cdots,a,\color{#0afd}{b'})\ $ if $ \bmod{\color{#0a0}a}\!:\ \color{#c00}b\equiv\color{#0af}{b'}\ \, $ [$\rm\color{darkorange}{MR}=$ gcd mod reduction]
$\!\begin{align} {\bf Lemma}\ &\ {\rm if}\, \bmod\ \color{#0a0}{\!f_{i}(n)}\!:\ \color{#c00}{f_{i}(n\!+\!1)}\equiv \color{#0af}{f_{i-1}(n)}\ \ {\rm for\ all}\,\ i,n\\[.4em] {\rm then}\,\ \ &\gcd(\color{0a0}{f_k(m)},\ \color{c00}{f_k(m\!+\!1)},\cdots,\color{0a0}{f_k(m\!+\!k\!-\!1)},\color{c00}{f_k(m\!+\!k)})\\[.4em] =\ &\gcd(f_k(m),\ f_{k-1}(m),\,\cdots,\, f_0(m))\,\ \ [=1\ \ {\rm if}\,\ f_0(m)=1] \end{align}$
$\begin{align}{\bf Proof}\!:\ &\gcd(\color{#0a0}{f_k(m)},\ \color{#c00}{f_k(m\!+\!1)},\cdots,\color{#0a0}{f_k(m\!+\!k\!-\!1)},\color{#c00}{f_k(m\!+\!k)})\\[.4em] =\ &\gcd(f_k(m),\underbrace{\color{#0af}{f_{k-1}(m)},f_{k-1}(m\!+\!1),\ \ \cdots\ \ ,\color{#0af}{f_{k-1}(m\!+\!k\!-\!1)}}_{\textstyle \gcd({f_{k-1}(m),\,\cdots,\,f_1(m), f_0(m))\ \ \rm by\ induction\!\!\!\!\!}}\!)\ \ {\rm by}\ \ \rm\color{darkorange}{MR}\\[.3em] \end{align}$
Bill Dubuque
- 282,220
-
OP is case $,f_i(n) = {n\choose i}\ $ so $,\color{#c00}{f_i(n!+!1)} = \color{0af}{f_{i-1}(n)}+ \color{0a0}{f_i(n)}\equiv \color{#0af}{f_{i-1}(n)}\pmod{!\color{#0a0}{f_i(n)}}\ \ $ – Bill Dubuque Jun 21 '24 at 19:52