Number Theory Archives - Epsilonify

If p divides a^2, then p divides a

The Mathematician — Thu, 17 Nov 2022 13:00:19 +0000

Let $p$ be a prime and $a$ an integer. If $p$ divides $a^2$ , then $p$ divides $a$

Proof. Given that

p

divides

a^2

, so that means that:

\begin{align*} p \mid a^2. \end{align*}

Each integer can be written as a unique prime factorization. Therefore:

\begin{align*} a = \prod_{i = 1}^{n} p_i^{m_i}. \end{align*}

This means that:

\begin{align*} p \mid (\prod_{i = 1}^{n} p_i^{m_i})^2 \iff p \mid \prod_{i = 1}^{n} p_i^{2m_i} \end{align*}

and

p = p_i

for some

i \in \{1,2,\ldots,n\}

p

is prime. Therefore:

\begin{align*} p \mid \prod_{i = 1}^{n} p_i^{m_i} \iff p \mid a, \end{align*}

which concludes the proof.

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If x = a (mod p) and x = a (mod q), then x = a (mod pq)

The Mathematician — Tue, 15 Nov 2022 13:00:30 +0000

Let $p$ and $q$ be distinct primes. If $x \equiv a \ (\text{mod } p)$ and $x \equiv a \ (\text{mod } q)$ , then $x \equiv a \ (\text{mod } pq)$ .

Proof. Since

x \equiv a \ (\text{mod } p)

and

x \equiv a \ (\text{mod } q)

, we have that

\begin{align*} p &\mid (x - a) \\ q &\mid (x - a). \end{align*}

By the fact that each integer has a unique prime factorization, we see that

\begin{align*} x - a = \sum_{i = 1}^l r_i^{m_i} \end{align*}

where

r_i

are distinct primes and

p = r_j

and

q = r_k

for some

i \in \{1,2,\ldots,l\}

. Therefore, we get

\begin{align*} pq \mid (x - a). \end{align*}

which implies that

x \equiv a \ (\text{mod } pq)

The post If x = a (mod p) and x = a (mod q), then x = a (mod pq) appeared first on Epsilonify.

If p divides n, then p-1 divides phi(n)

The Mathematician — Tue, 01 Nov 2022 13:00:48 +0000

Let $p$ be a prime number. If $p \mid n$ , then $p - 1 \mid \phi(n)$ .

Proof. We need to prove that if

p

divides

n

, then

p-1

divides the Euler’s totient function of

n

. Assume that

p \mid n

. Then

n = p^{\alpha}c

for some integer

c

and positive integer

\alpha

where

p

doesn’t divide

c

. Now since

p \nmid c

, this implies that

gcd(p^{\alpha},c) = 1

. Since the Euler’s totient function is multiplicative, we can apply that

\phi(p^{\alpha}c) = \phi(p^{\alpha})\phi(c)

. So together, we get

\begin{align*} \phi(n) &= \phi(p^{\alpha}c) \quad \quad \text{ since } n = p^{\alpha}c \\ &=\phi(p^m)\phi(c) \quad \text{ since } gcd(p^{\alpha},c) = 1 \\ &= p^{\alpha - 1}(p-1)\phi(c). \end{align*}

So since

\phi(n) = p^{\alpha - 1}(p-1)\phi(c)

, we have that

p - 1 \mid \phi(n)

The post If p divides n, then p-1 divides phi(n) appeared first on Epsilonify.

If a,b is in (Z/nZ), then ab is in (Z/nZ)

The Mathematician — Sun, 30 Oct 2022 13:00:07 +0000

If $a,b \in (\mathbb{Z}/n\mathbb{Z})^{\times}$ , then $a,b \in (\mathbb{Z}/n\mathbb{Z})^{\times}$ .

Proof. By definition, if

a,b \in (\mathbb{Z}/n\mathbb{Z})^{\times}

, then

\begin{align*} ac &= 1 \in \mathbb{Z}/n\mathbb{Z} \text{ for some } c \in \mathbb{Z}/n\mathbb{Z} \\ bd &= 1 \in \mathbb{Z}/n\mathbb{Z} \text{ for some } d \in \mathbb{Z}/n\mathbb{Z}. \end{align*}

Obviously, we will get

\begin{align*} 1 = 1 \cdot 1 = ac \cdot bd = ab \cdot cd, cd \in \mathbb{Z}/n\mathbb{Z}, \end{align*}

which implies that

ab \in (\mathbb{Z}/n\mathbb{Z})^{\times}

The post If a,b is in (Z/nZ)*, then ab is in (Z/nZ)* appeared first on Epsilonify.