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Reducing $(458, 214, 25)$ to $(1, 0, 1)$

Example

The positive-definite form $(458, 214, 25)$ reduces to $(1, 0, 1)$ through the explicit swap-and-shear moves

$(458, 214, 25) \to (25, - 214, 458) \to (25, - 14, 2) \to (2, 14, 25) \to (2, 2, 1) \to (1, - 2, 2) \to (1, 0, 1) .$

Equivalently,

$(458, 214, 25) ∣ (\begin{array}{cc} 3 & 4 \\ - 13 & - 17 \end{array}) = (1, 0, 1) .$

Facts & Assumptions

Given: The integral form $f = (458, 214, 25)$ .

[L1]

Integral substitution defines a right action of ${S L}_{2} (Z)$ on integral binary quadratic forms (Integral substitution defines a right action of ${S L}_{2} (Z)$ on integral binary quadratic forms).

[L2]

Every positive-definite integral binary quadratic form is properly equivalent to a reduced form (Every positive-definite integral binary quadratic form is properly equivalent to a reduced form).

[L3]

A positive-definite form is reduced when $∣ b ∣ \leq a \leq c$ and the boundary sign condition holds (Reduced positive-definite binary quadratic forms).

Verification

technique · direct

1.1L1givenalgebra

Let $S = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$ and $T_{k} = (\begin{array}{cc} 1 & k \\ 0 & 1 \end{array})$ . Direct substitution gives $f ∣ S = (25, - 214, 458)$ , $(f ∣ S) ∣ T_{4} = (25, - 14, 2)$ , $((f ∣ S) ∣ T_{4}) ∣ S = (2, 14, 25)$ , then after $T_{- 3}$ one gets $(2, 2, 1)$ , after another $S$ one gets $(1, - 2, 2)$ , and after $T_{1}$ one gets $(1, 0, 1)$ .

2.1L3step 1.1algebra

The final form $(1, 0, 1)$ is reduced because $∣ 0 ∣ \leq 1 \leq 1$ and the boundary sign condition is automatic.

2.2L1step 1.1algebra

By repeated use of the right-action law [L1], the composite matrix is $S T_{4} S T_{- 3} S T_{1} = (\begin{array}{cc} 3 & 4 \\ - 13 & - 17 \end{array})$ , which has determinant $1$ , so the single displayed substitution is exactly the product of the six moves in step 1.1.

3.1L2step 2.1step 2.2∎

Thus the explicit reduction algorithm indeed carries $(458, 214, 25)$ to the reduced form $(1, 0, 1)$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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

William Stein, Elementary Number Theory and Elliptic Curves, Examples 9.2.5 and 9.3.3 (standard reference, not scraped)

Reducing (458,214,25)(458,214,25) to (1,0,1)(1,0,1)