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Prime factorisation gives two representations of $221$ as a sum of two squares

Example

The factorisation $221 = 13 \cdot 17$ and the two sign variants in the two-square identity give

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

Facts & Assumptions

Given: The integers $13$ , $17$ , and their product $221$ .

[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 all integers $a, b, c, d$ , $(a^{2} + b^{2}) (c^{2} + d^{2}) = (a c - b d)^{2} + (a d + b c)^{2} = (a c + b d)^{2} + (a d - b c)^{2}$ (The Brahmagupta–Fibonacci two-square identity).

Verification

technique · direct

1.1

Directly, $13 = 2^{2} + 3^{2}$ and $17 = 1^{2} + 4^{2}$ .

F1algebra

2.1

The first sign variant in [L1] gives $221 = (- 10)^{2} + 11^{2} = 10^{2} + 11^{2}$ , since $2 \cdot 1 - 3 \cdot 4 = - 10$ and $2 \cdot 4 + 3 \cdot 1 = 11$ .

step 1.1L1algebra

2.2

The second sign variant gives $221 = 14^{2} + 5^{2}$ , since $2 \cdot 1 + 3 \cdot 4 = 14$ and $2 \cdot 4 - 3 \cdot 1 = 5$ .

step 1.1L1algebra

3.1

Both pairs satisfy [F1]. They are essentially different because the unordered absolute-coordinate sets ${10, 11}$ and ${5, 14}$ differ.

step 2.1step 2.2F1algebra∎

Depends on

Used by

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Sources

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

Prime factorisation gives two representations of 221221 as a sum of two squares