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回复 1# nttz 本thread可以分类“数论”
下面证明这个因式分解的系数$\gcd\left(\binom{n}{1} , \binom{n}{2} , \binom{n}{3},\ldots,\binom{n}{\lfloor \frac n2 \rfloor}\right)$必为1或素数.
搬运自math.stackexchange.com/questions/2067235/gcd- … inomial-coefficients
$\newcommand{\Z}{\mathbb{Z}}$Suppose $n$ is not a prime-power, and let
$$n = \prod_{i=1}^{k} p_{i}^{e_{i}}$$
with $p_{i}$ distinct primes, $e_{i} > 0$ for each $i$.
Write
$$n_{j} = \prod_{i \ne j} p_{i}^{e_{i}}.$$
Then it is well known that
$$
\binom{n}{p_{i}^{e_{i}}} \equiv \binom{p_{i}^{e_{i}} n_{i}}{p_{i}^{e_{i}}} \equiv n_{i} \pmod{p},
$$
so that $\dbinom{n}{p_{i}^{e_{i}}}$ is not divisible by $p_{i}$. Since your gcd is a divisor of $n = \dbinom{n}{1}$, this shows that the gcd is $1$ in this case.
If $n = p^{e}$ is a power of the prime $p$, then the equation in $\Z/p\Z[x]$
$$
(1 + x)^{p^{e}} = 1 + x^{p^{e}}
$$
shows that all of your binomial coefficients are divisible by $p$, and their gcd divides $p^e = \dbinom{p^e} {1}$.
But by Kummer, the highest power of $p$ dividing
$$
\binom{p^{e}}{p^{e-1}}
$$
is $p$, as there is precisely one carry in the $p$-adic addition of $p^{e-1}$ and $p^{e} - p^{e-1} = (p-1) p^{e-1}$, so that your gcd is $p$.
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kuing
Posted 2022-5-1 19:09
