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83 lines
3.2 KiB
Markdown
83 lines
3.2 KiB
Markdown
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# Handy Quadratic Congruences Facts
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## Number of Solutions
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<u>For congruences mod 2</u>
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**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,
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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
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2. If $\Delta \equiv 5$ (mod $8$), so that $b$ and $c$ are odd, then $f(x) \equiv 0$ (mod $2$) has **no** solutions
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3. If $4 | \Delta$ , so that $b$ is even, then $f(x) \equiv 0$ (mod $2$) has exactly **one** solution.
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**Proposition 16.2.** Let $p$ be an odd prime and let $a$ be an integer. Then,
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1. If $p$ does not divide $a$, then the congruence $x^2 \equiv a$ (mod $p$) has either two solutions or no solutions.
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2. If $p$ divides $a$, then $x^2 \equiv a$ (mod $p$) has exactly one solution, namely $x = 0$.
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**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
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$$
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(\frac{a}{p}) = \begin{cases}
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1, & \text{if $x^2 \equiv a$ (mod $p$) has exactly two solutions,} \\
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0, & \text{if $x^2 \equiv a$ (mod $p$) has exactly one solution,} \\
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-1, & \text{if $x^2 \equiv a$ (mod $p$) has no soultions.}
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\end{cases}
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$$
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**Properties of Legendre symbol**.
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- $(\frac{a}{p}) = 0 \iff p$ divides $a$
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- $(\frac{1}{p}) = 1$ for every odd prime $p$
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- $a \equiv b$ (mod $p$) $\implies$ $(\frac{a}{p}) = (\frac{b}{p})$
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- $(\frac{ab}{p}) = (\frac{a}{p})(\frac{b}{p})$
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- If $p$ is an odd prime then,
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- $$
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(\frac{-1}{p}) =
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\begin{cases}
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1, & \text{if $p \equiv 1$ (mod $4$)} \\
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-1, & \text{ if $p \equiv 3$ (mod $4$)}
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\end{cases}
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$$
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- If $p$ is an odd prime then,
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- $$
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(\frac{2}{p}) =
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\begin{cases}
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1, & \text{if $p \equiv 1$ (mod $8$) or $p \equiv 7$ (mod $8$)} \\
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-1, & \text{ if $p \equiv 3$ (mod $8$) or $p \equiv 5$ (mod $8$)}
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\end{cases}
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$$
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- **Quadratic Reciprocity Theorem**. Let $p$ and $q$ be distinct odd primes. Then,
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- $$
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(\frac{q}{p}) =
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\begin{cases}
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(\frac{p}{q}), & \text{if $p \equiv 1$ (mod $4$) or $q \equiv 1$ (mod $4$)} \\
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-(\frac{p}{q}), & \text{ if $p \equiv 3$ (mod $4$) and $q \equiv 3$ (mod $4$)}
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\end{cases}
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$$
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## Procedure
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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.
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\begin{align*}
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ax^2 + bx + c \equiv 0 \text{ (mod $p$)} &\iff 4a(ax^2 + bx + c) \equiv 0 \text{ (mod $p$)} \\\\
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&\iff 4a^2x^2 + 4abx + 4ac \equiv 0 \text{ (mod $p$)} \\\\
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&\iff 4a^2x^2 + 4abx + 4ac + (b^2 - 4ac) \equiv b^2 - 4ac \text{ (mod $p$)} \\\\
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&\iff 4a^2x^2 + 4abx + b^2 \equiv b^2 - 4ac \text{ (mod $p$)} \\\\
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&\iff (2ax+b)^2 \equiv b^2 - 4ac \text{ (mod $p$)}
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\end{align*}
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## Quadratic Congruences Modulo a prime power
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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 $p$).
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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 $p$)
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