## Brandon Rozek

Software Developer, Researcher, and Linux Enthusiast.

## Number of Solutions

For congruences mod 2

Proposition 16.1. Let $f(x) = ax^2 + bx + c$ with $a$ odd, and let $\Delta = b^2 - 4ac$ be the discriminant of $f(x)$. Then,

1. If $\Delta \equiv 1$ (mod $8$), so that $b$ is odd and $c$ is even, then $f(x) \equiv 0$ (mod $2$) has two solutions
2. If $\Delta \equiv 5$ (mod $8$), so that $b$ and $c$ are odd, then $f(x) \equiv 0$ (mod $2$) has no solutions
3. If $4 | \Delta$ , so that $b$ is even, then $f(x) \equiv 0$ (mod $2$) has exactly one solution.

Proposition 16.2. Let $p$ be an odd prime and let $a$ be an integer. Then,

1. If $p$ does not divide $a$, then the congruence $x^2 \equiv a$ (mod $p$) has either two solutions or no solutions.
2. If $p$ divides $a$, then $x^2 \equiv a$ (mod $p$) has exactly one solution, namely $x = 0$.

Legendre symbol definition. Let $p$ be an odd prime and $a$ any integer. Then the Legendre symbol, written as $(\frac{a}{p})$ is defined as $$(\frac{a}{p}) = \begin{cases} 1, & \text{if x^2 \equiv a (mod p) has exactly two solutions,} \ 0, & \text{if x^2 \equiv a (mod p) has exactly one solution,} \ -1, & \text{if x^2 \equiv a (mod p) has no soultions.} \end{cases}$$ Properties of Legendre symbol.

• $(\frac{a}{p}) = 0 \iff p$ divides $a$

• $(\frac{1}{p}) = 1$ for every odd prime $p$

• $a \equiv b$ (mod $p$) $\implies$ $(\frac{a}{p}) = (\frac{b}{p})$

• $(\frac{ab}{p}) = (\frac{a}{p})(\frac{b}{p})$

• If $p$ is an odd prime then,

• $$(\frac{-1}{p}) = \begin{cases} 1, & \text{if p \equiv 1 (mod 4)} \ -1, & \text{ if p \equiv 3 (mod 4)} \end{cases}$$
• If $p$ is an odd prime then,

• $$(\frac{2}{p}) = \begin{cases} 1, & \text{if p \equiv 1 (mod 8) or p \equiv 7 (mod 8)} \ -1, & \text{ if p \equiv 3 (mod 8) or p \equiv 5 (mod 8)} \end{cases}$$
• Quadratic Reciprocity Theorem. Let $p$ and $q$ be distinct odd primes. Then,

• $$(\frac{q}{p}) = \begin{cases} (\frac{p}{q}), & \text{if p \equiv 1 (mod 4) or q \equiv 1 (mod 4)} \ -(\frac{p}{q}), & \text{ if p \equiv 3 (mod 4) and q \equiv 3 (mod 4)} \end{cases}$$

## Procedure

When $p$ is an odd prime, a quadratic congruence $ax^2 + bx + c \equiv 0$ (mod $p$) can be transformed into a specialized form by completing the square. \begin{align*} ax^2 + bx + c \equiv 0 \text{ (mod $p$)} &\iff 4a(ax^2 + bx + c) \equiv 0 \text{ (mod $p$)} \\ &\iff 4a^2x^2 + 4abx + 4ac \equiv 0 \text{ (mod $p$)} \\ &\iff 4a^2x^2 + 4abx + 4ac + (b^2 - 4ac) \equiv b^2 - 4ac \text{ (mod $p$)} \\ &\iff 4a^2x^2 + 4abx + b^2 \equiv b^2 - 4ac \text{ (mod $p$)} \\ &\iff (2ax+b)^2 \equiv b^2 - 4ac \text{ (mod $p$)} \end{align*}

## Quadratic Congruences Modulo a prime power

Let $a$ be the solution to $f(x) \equiv 0$ (mod $p$) where $p$ is an odd prime. Consider $b = pt + a$. Then, $f(b) \equiv 0$ (mod $p^2$) if $f^\prime(a)t \equiv -\frac{f(a)}{p}$ (mod $p$).

In general, let $a$ be the solution to $f(x) \equiv 0$ (mod $p^n$) where $p$ is an odd prime. Consider $b = pt + a$. Then, $f(b) \equiv 0$ (mod $p^{n + 1}$) if $f^\prime(a)t \equiv -\frac{f(a)}{p^n}$ (mod $p$)