Proof of Some Square Patterns

$(\frac{- 1}{p}) = 1 ⟺ p \equiv 1 mod 4$ by supplementary law for $(\frac{- 1}{p})$ . $(\frac{p}{5}) = 1 ⟺ p \equiv 1, 4 mod 5$ .

p mod 4 11 p mod 5 14 p mod 20 19

$(\frac{- 1}{p}) = - 1 ⟺ p \equiv 3 mod 4$ by supplementary law for $(\frac{- 1}{p})$ . $(\frac{p}{5}) = - 1 ⟺ p \equiv 2, 3 mod 5$ .

p mod 4 33 p mod 5 23 p mod 20 73

Note:

(\frac{6}{p}) = (\frac{2}{p}) (\frac{3}{p}) = (\frac{2}{p}) ((- 1)^{(\frac{p - 1}{2}) (\frac{3 - 1}{2})} (\frac{p}{3})) by Main Law = (\frac{2}{p}) (- 1)^{\frac{p - 1}{2}} (\frac{p}{3}) \to 3 Legendre symbols depending on p mod 8, and p mod 3, which is p mod 24.

On Problem Set 1 or 2 we checked each $a mod p$ is $x^{2} + y^{2} mod p$ ( $p = 2, \dots, 29$ ).

Show: Each $a mod p$ is $x^{2} + y^{2} mod p$ for every prime $p$ by giving a formula for $# {x^{2} + y^{2} \equiv a mod p}$ .

N_{n, p} = # {(x_{1}, \dots, x_{n}) mod p : x_{1}^{2} + x_{2}^{2} + \dots + x_{n}^{2} \equiv 1 mod p}

Find a recursion for $N_{n,p}$ from $N_{n-2,p}$ for $n \geq 3$ by induction on $n$

For distinct primes $p, q > 2$ , we’ll calculate $N_{q, p} mod p$ in two ways and the quadratic reciprocity will fall out.

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