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Odd primes congruent to $1$ or $3$ modulo $8$ are represented by $x^{2} + 2 y^{2}$

Example

An odd prime $p$ is represented by $x^{2} + 2 y^{2}$ if and only if

$p \equiv 1 (m o d 8) or p \equiv 3 (m o d 8) .$

Facts & Assumptions

Given: An odd prime $p$ .

[L1]

The unique reduced primitive form of discriminant $- 8$ is $(1, 0, 2)$ , so every primitive positive-definite form of discriminant $- 8$ is properly equivalent to $x^{2} + 2 y^{2}$ (The reduced primitive forms of discriminant $- 8$ ).

[L2]

Properly equivalent binary quadratic forms represent the same integers (Properly equivalent binary quadratic forms represent the same integers, with primitive representations in bijection).

[L3]

An integral binary quadratic form is positive definite exactly when its leading coefficient is positive and its discriminant is negative (An integral binary quadratic form is positive definite exactly when its leading coefficient is positive and its discriminant is negative).

[L4]

For an odd prime $p$ , $(\frac{- 1}{p}) = 1$ exactly when $p \equiv 1 (m o d 4)$ , and $(\frac{- 1}{p}) = - 1$ exactly when $p \equiv 3 (m o d 4)$ (First supplement: $(- 1 / p) = (- 1)^{(p - 1) / 2}$ ).

[L5]

For an odd prime $p$ , $(\frac{2}{p}) = 1$ exactly when $p \equiv 1$ or $7 (m o d 8)$ , and $(\frac{2}{p}) = - 1$ exactly when $p \equiv 3$ or $5 (m o d 8)$ (Second supplement: $(2 / p) = (- 1)^{(p^{2} - 1) / 8}$ ).

[L6]

For every odd prime $p$ and integers $u, v$ , $(\frac{u v}{p}) = (\frac{u}{p}) (\frac{v}{p})$ (The Legendre symbol is multiplicative for all integer numerators).

[L7]

For an odd prime $p$ , $(\frac{a}{p}) = 1$ exactly when $a$ is a quadratic residue modulo $p$ (The Legendre symbol, including its zero value).

Verification

technique · direct

1.1

If $p = x^{2} + 2 y^{2}$ , then reducing modulo $8$ shows $x^{2} \equiv 1 (m o d 8)$ and $2 y^{2} \equiv 0$ or $2 (m o d 8)$ , because squares modulo $8$ are $0, 1, 4$ and $p$ is odd. Hence $p \equiv 1$ or $3 (m o d 8)$ .

givenalgebra

1.2

Conversely, suppose $p \equiv 1$ or $3 (m o d 8)$ . Then [L4] and [L5] give $(\frac{- 1}{p}) (\frac{2}{p}) = 1$ , so [L6] yields $(\frac{- 2}{p}) = 1$ . By [L7], there is an integer $t$ with $t^{2} \equiv - 2 (m o d p)$ .

L4L5L6L7givenalgebra

2.1

Set $c = (t^{2} + 2) / p$ , which is an integer by step 1.2, and define $f = (p, 2 t, c) .$ Then $Δ (f) = (2 t)^{2} - 4 p c = 4 t^{2} - 4 (t^{2} + 2) = - 8,$ and $f (1, 0) = p$ , so $f$ represents $p$ . If a divisor $d$ of $p$ , $2 t$ , and $c$ were greater than $1$ , then $d = p$ because $p$ is prime; but $p ∣ 2 t$ would force $p ∣ t$ , hence $p ∣ t^{2} + 2$ , so $p ∣ 2$ , impossible for an odd prime. Thus $f$ is primitive. Since its leading coefficient is $p > 0$ and its discriminant is $- 8 < 0$ , [L3] makes $f$ positive definite. Therefore [L1] shows that $f$ is properly equivalent to $x^{2} + 2 y^{2}$ , and then [L2] shows that $x^{2} + 2 y^{2}$ represents $p$ .

L1L2L3step 1.2givenconstructalgebra

3.1

Steps 1.1 and 2.1 prove the claimed criterion.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Andrew Granville, Binary Quadratic Forms, Exercise 4.1h (standard reference, not scraped)

Odd primes congruent to 11 or 33 modulo 88 are represented by x2+2y2x2+2y2