p-adic分析note| Solving Congruences Modulo $p^n$

p-adic分析note| Solving Congruences Modulo $p^n$
p-adic分析note| Solving Congruences Modulo $p^n$

Solving Congruences Modulo $p^n$

The “p-adic numbers” we have just constructed are closely related to the problem of solving congruences modulo powers of $p$. We will look at some examples of this.

Let’s start with the easiest possible case, an equation which has solutions in $\mathbb{Q}$, such as
$$
X^{2}=25 .
$$
We want to consider it modulo $p^{n}$ for every $n$, i.e., to solve the congruences
$$
X^{2} \equiv 25 \quad\left(\bmod p^{n}\right) .
$$
Now, of course, our equation has solutions already in the integers: $X=$ $\pm 5$. This automatically gives solutions of the congruence for every $n$; just put $X \equiv \pm 5\left(\bmod p^{n}\right)$ for every $n$.

p-adic分析note| Solving Congruences Modulo $p^n$

解同余模 $p^{n}$


我们刚刚构造的“p进数”与解决同余模幂的问题密切相关 $p$. 我们将看一些这样的例子。
让我们从最简单的情况开始,一个有解的方程 $\mathbb{Q}$ ,如
$$
X^{2}=25 \text {. }
$$
我们想考虑它模 $p^{n}$ 对于每个 $n$ ,即解决同余
$$
X^{2} \equiv 25 \quad\left(\bmod p^{n}\right) .
$$
现在,当然,我们的方程已经有了整数解: $X=\pm 5$. 这会自动为每个 $n$; 就放 $X \equiv \pm 5\left(\bmod p^{n}\right)$ 对于每个 $n$.

p-adic分析note| Solving Congruences Modulo $p^n$
p-adic分析note| Solving Congruences Modulo $p^n$

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