binomial coefficient | 我不是蜜斯佛陀

題目：設 $p$ 是質數， $a, b$ 是整數，如果 $\displaystyle\frac{a^p-b^p}{p}$ 是整數，證明 $\displaystyle\frac{a^p-b^p}{p^2}$ 也是整數。

解答：Suppose we have know the following lemma

If $p$ is prime, then ${ p \choose k}$ is divisible by $p$ if and only if nonnegative integer $k\neq 0, p$

For convenience, let $\displaystyle\frac{a^p-b^p}{p} = n$ and $a = b+c$ , where $c$ is also an integer.

Then we can rewrite $n$ as

$\displaystyle\frac{(b+c)^p-b^p}{p} = \frac{\sum_{k=1}^{p-1}{ p \choose k}c^{k}b^{p-k} + c^p}{p}$

Since $n$ is an integer, then $c^p$ is divisible by $p$ .

Thus $c$ is also divisible by $p$ , since $p$ is prime.

Then ${ p \choose k}c^{k}b^{p-k}$ is divisible by $p^2$ , $\forall 1\leq k < p$

and $c^p$ is also divisible by $p^2$ since $p \geq 2$ .

Hence $n$ is divisible by $p$ , that means $\displaystyle\frac{a^p-b^p}{p^2}$ is an integer

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