# 数学代写|数论代写number theory代考 M328K 55385|Congruences

#### Doug I. Jones

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## 数学代写|数论代写number theory代考|Basic congruences

If $m, b$, and $c$ are integers for which $m$ divides $b-c$, then we write
$$b \equiv c \quad(\bmod m)$$
and say that $b$ and $c$ are congruent modulo $m$, where $m$ is the modulus 1 The numbers involved should be integers, not fractions, and the modulus can be taken in absolute value; that is, $b \equiv c(\bmod m)$ if and only if $b \equiv c(\bmod |m|)$, by definition.

For example, $-10 \equiv 15(\bmod 5)$, and $-7 \equiv 15(\bmod 11)$, but $-7 \not \equiv 15$ $(\bmod 3)$. Note that $b \equiv b(\bmod m)$ for all integers $m$ and $b$.

## 数学代写|数论代写number theory代考|The trouble with division

Although the rules for addition, subtraction, and multiplication work for congruences as they do for the integers, reals, and most other mathematical objects we have encountered, the rule for division is more subtle. In the complex numbers, if we are given numbers $a$ and $b \neq 0$, then there exists a unique value of $c$ for which $a=b c$ (so that $c=a / b$ ), and therefore there is no ambiguity in the definition of division. We now look at the multiplication tables mod 5 and mod 6 to see whether this same property holds for modular arithmetic:

## 数学代写|数论代写number theory代考|Congruences for polynomials

Let $\mathbb{Z}[x]$ denote the set of polynomials with integer coefficients. Using the above rules for congruences, one gets a very useful result for congruences involving polynomials:

Corollary 2.3.1. If $f(x) \in \mathbb{Z}[x]$ and $a \equiv b(\bmod m)$, then $f(a) \equiv f(b)(\bmod m)$.
Proof. Since $a \equiv b(\bmod m)$ we have $a^{2} \equiv b^{2}(\bmod m)$ by Lemma 2.1.1, and then
Exercise 2.3.1. Prove that $a^{k} \equiv b^{k}(\bmod m)$ for all integers $k \geq 1$, by induction.
Now, writing $f(x)=\sum_{i=0}^{d} f_{i} x^{i}$ where each $f_{i}$ is an integer, we have
$$f(a)=\sum_{i=0}^{d} f_{i} a^{i} \equiv \sum_{i=0}^{d} f_{i} b^{i}=f(b) \quad(\bmod m),$$
by Lemma $2.1 .1$
This result can be extended to polynomials in many variables.
Exercise 2.3.2. Deduce, from Corollary [2.3.1 that if $f(t) \in \mathbb{Z}[t]$ and $r, s \in \mathbb{Z}$, then $r-s$ divides $f(r)-f(s)$

Therefore, for any polynomial $f(x) \in \mathbb{Z}[x]$, the sequence $f(0), f(1), f(2), \ldots$ modulo $m$ is periodic of period $m$; that is, the values repeat every $m$ th term in the sequence, repeating indefinitely. More precisely $f(n+m) \equiv f(n)(\bmod m)$ for all integers $n$.

## 数学代写|数论代写number theory代考|Basic congruences

$$b \equiv c \quad(\bmod m)$$

## 数学代写|数论代写number theory代考|Congruences for polynomials

$$f(a)=\sum_{i=0}^{d} f_{i} a^{i} \equiv \sum_{i=0}^{d} f_{i} b^{i}=f(b) \quad(\bmod m),$$

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