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Number Theory Notes

Wilson's Theorem

$p$ prime

$$ \prod_{i=1}^{p-1} (i) \mod p \equiv -1 \mod p $$

proof:

$$ \begin{equation} \label{eq1} \begin{split} \prod_{i=1}^{p-1} (i) & = 1 \cdot 2 \cdot 3 \cdots (p-1) \\ & = (a_0 a_0') (a_1 a_1^{-1}) \cdots (a_{\frac{p-1}{2}} a_{\frac{p-1}{2}}^{-1}) \\ & = [1] \end{split} \end{equation} $$

Where $a_0=1$ and $a_0' = -1$.

lemma:

The only number whose inverse is itself is $(-1)$

lemma proof:

$$ \begin{split} x^2 & = 1 \mod p \\ & = \pm 1 \mod p \end{split} $$

Since $p$ is prime, there are only two solutions.

proof (cont'd):

$$ \begin{split} \to [1] & = (1 \cdot -1) (a_1 a_1^{-1}) (a_2 a_2^{-1}) \cdots (a_{\frac{p-1}{2}} a_{\frac{p-1}{2}}^{-1}) \\ & = (1 \cdot -1) (1) (1) \cdots (1) \\ & = -1 \end{split} $$


Fermat's Theorem

$p$ prime

$$ \begin{split} x^{p-1} = 1 \mod p \end{split} $$

proof:

$a \ne 0$

$$ \begin{align} & & -1 & = 1 \cdot 2 \cdot 3 \cdot 4 \cdots (p-1) \mod p \\ \to & & -a^{p-1} & = (a \cdot 1) (a \cdot 2) (a \cdot 3) \cdots (a \cdot (p-1)) \mod p \\ \to & & -a^{p-1} & = 1 \cdot 2 \cdot 3 \cdot 4 \cdots (p-1) \mod p \\ \to & & -a^{p-1} & = -1 \mod p \\ \to & & a^{p-1} & = 1 \mod p \end{align} $$

Since $a \cdot x$ is 1-1 and onto for prime $p$ (with $a,x \ne 0$).

Legendre Symbol

$p$ prime

$$ \left( \dfrac{a}{p} \right) = \begin{cases} 1, & \text{ if } \sqrt{a} \text{ exists } \\ 0, & \text{ if } \gcd(a,p) \ne 1 \\ -1, & \text{ if } \sqrt{a} \text{ does not exist } \end{cases} $$

notes:

$g$ a generator of $p$, consider $a = g^\beta$

$\beta$ even:

$$ \begin{align} & & g^\beta & = a \\ \to & & ( g^{\frac{p-1}{2}} ) )^\beta & = ( a^{\frac{p-1}{a}} ) \\ \to & & (-1)^\beta & = a^{\frac{p-1}{2}} \\ \to & & a^{\frac{p-1}{2}} & = 1 \\ \end{align} $$

$\beta$ odd:

$$ \begin{align} \to & & (-1)^\beta & = a^{\frac{p-1}{2}} \\ \to & & a^{\frac{p-1}{2}} & = -1 \\ \end{align} $$

So:

$$ \left( \dfrac{a}{p} \right) = a^{\frac{p-1}{2}} $$


Zeta Function

$$ \begin{split} \sum_{n=1}^{\infty} \frac{1}{n^s} & = \prod_{p \text{ prime}}^{\infty} ( \frac{1}{1 - p^{-s}} ) \end{split} $$

proof:

$$ \begin{align} & & \sum_{i=1}^{\infty} a^i & = S \\ & & \sum_{i=0}^{\infty} a^{i+1} & = \sum_{i=1}^{\infty} a^i \\ & & & = S a \\ & & & \\ & & S - S a & = 1 \\ \to & & S & = \frac{1}{1-a} \end{align} $$

Write out the product of infinit series of primes, $p_i$:

$$ \begin{align} & ( 1 + p_0^{-s} + p_0^{-2s} + p_0^{-3s} + \cdots ) \\ & ( 1 + p_1^{-s} + p_1^{-2s} + p_1^{-3s} + \cdots ) \\ & \cdots \\ & = \prod_{p \text{ prime}}^{\infty} [ \sum_{i=0}^{\infty} \frac{1}{p^{i s}} ] \\ & = \prod_{p \text{ prime}}^{\infty} ( \frac{1}{1-p^{-s}} ) \end{align} $$

Any choice of terms in the product of infinite sums of primes will yield a value in the original $\sum \frac{1}{n^s}$.


2019-01-10