website/content/notes/quadraticcongruences.md

Handy Quadratic Congruences Facts

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)