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The representations $221 = 5^{2} + 14^{2} = 11^{2} + 10^{2}$ recover the factors $13$ and $17$

Example

The normalized representations

$221 = 5^{2} + 14^{2} = 11^{2} + 10^{2}$

feed the factorisation construction and recover $221 = 13 \cdot 17$ .

Facts & Assumptions

Given: The displayed representations of $221$ .

[F1]

A representation of a nonnegative integer $n$ as a sum of two squares is an ordered pair $(x, y) \in Z^{2}$ such that $n = x^{2} + y^{2}$ (Representations and primitive representations as sums of two squares).

[L1]

For an odd integer, two essentially different normalized representations force a factorisation $N = P Q$ with $P, Q > 1$ (Two essentially different two-square representations factor an odd integer).

Verification

technique · direct

1.1F1algebra

Both equalities are direct, and $(5, 14)$ and $(11, 10)$ are positive odd-even normalized pairs with $0 < 5 < 11$ .

2.1step 1.1L1algebra

In the notation of [L1], the values $e = 2$ , $f = 3$ , $g = 4$ , $h = 1$ give $e g - f h = 5$ , $f g + e h = 14$ , $e g + f h = 11$ , and $f g - e h = 10$ .

3.1step 2.1F1algebra∎

The resulting factors are $e^{2} + f^{2} = 13$ and $g^{2} + h^{2} = 17$ , and $13 \cdot 17 = 221$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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

P. Hackman, Elementary Number Theory, Chapter E, Theorem E.I.3 and Example E.II.3(a) (standard reference, not scraped)

The representations 221=52+142=112+102221=52+142=112+102 recover the factors 1313 and 1717