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Number Theory

This section covers number theory, especially modulos (moduli?).

Definitions

$n\equiv a\pmod{b}$ if $n$ is the remainder when $a$ is divided by $b$ to give an integral amount. Also, this means b divides (n-a).
$a|b$ (or $a$ divides $b$ ) if $b=ka$ for some integer $k$ .

Special Notation

Occasionally, if two equivalent expressions are both modulated by the same number, the entire equation will be followed by the modulo.

$(a_1, a_2,...a_n)$ refers to the greatest common factor of $a_1, a_2, ...a_n$ and $[a_1, a_2, ...a_n]$ refers to the lowest common multiple of $a_1, a_2,...a_n$ .

Properties

For any number there will be only one congruent number modulo $m$ between $0$ and $m-1$ .

If $a\equiv b \pmod{m}$ and $c \equiv d \pmod{m}$ , then $(a+c) \equiv (b+d) \pmod {m}$ .

$a \pmod{m} + b \pmod{m} \equiv (a + b) \pmod{m}$

$a \pmod{m} - b \pmod{m} \equiv (a - b) \pmod{m}$

$a \pmod{m} \cdot b \pmod{m} \equiv (a \cdot b) \pmod{m}$

Fermat's Little Theorem

For a prime $p$ and a number $a$ such that $(a,b)=1$ , $a^{p-1}\equiv 1 \pmod{p}$ . A frequently used result of this is $a^p\equiv a\pmod{p}$ .

Example: Find all primes p such that $p|2^p+1$ .

Solution: Firstly, p=2 clearly does not work. Now, as all other primes are odd, $(2, p)=1$ and hence $2^p\equiv2\pmod{p}$ . After adding one, we have $3\equiv0\pmod{p}$ since p divides $2^p+1$ . However, that means p must divide 3, so the only prime possible is 3. Indeed, $2^3+1=9$ is a multiple of 3.