Talk:Euler's factorization method

I have my own variation on the theme, which I shall demonstrate using the same numbers as in the worked example:

1000009 = 1000^2 + 3^2 = 972^2 + 235^2.

Pair off the squared numbers, odd with odd and even with even: {1000,972} and {235,3}.

Take one pair and put their half-sum and half-difference along the diagonal of a 2x2 square:

986 ===
=== 14

Fill in the remaining spaces with the half-sum and half-difference from the other pair:

986 119
116 14

Now calculate the ratios reading across and down:

986/119 = 116/14 = 58/7
986/116 = 119/14 = 17/2

986 119 17
116 14 2
58 7

And the factors are:
58^2 + 7^2 = 3413
17^2 + 2^2 = 293

86.4.253.180 (talk) 00:17, 12 June 2013 (UTC) 86.4.253.180 (talk) 00:21, 12 June 2013 (UTC) 86.4.253.180 (talk) 00:24, 12 June 2013 (UTC) [reply ]

"which apparently was previously thought to be prime even though it is not a pseudoprime by any major primality test." This sentence doesn't make sense. Typo maybe? — Preceding unsigned comment added by 50.46.174.233 (talk) 03:25, 7 December 2013 (UTC) [reply ]

Why doesn't this make any sense? Pieater3.14159265 (talk) 03:10, 30 July 2015 (UTC) [reply ]

Euler Can Factor From Two Equations Of a^2+D*y^2, not just from x^2+y^2

Euler seems to have another factoring method, mentioned in p362 of vol 1 of Dickson's "History of Numbers"^[1] :

Euler^[2] noted that ${\displaystyle N=a^{2}+\lambda *b^{2}=x^{2}+\lambda *y^{2}}${\displaystyle N=a^{2}+\lambda *b^{2}=x^{2}+\lambda *y^{2}} imply

${\displaystyle N=(1/4)*(\lambda *m^{2}+n^{2})*(\lambda *p^{2}+q^{2})}${\displaystyle N=(1/4)*(\lambda *m^{2}+n^{2})*(\lambda *p^{2}+q^{2})}, ${\displaystyle a\pm x=\lambda *m*p,n*q}${\displaystyle a\pm x=\lambda *m*p,n*q}, ${\displaystyle y\pm b=m*q,n*p}${\displaystyle y\pm b=m*q,n*p}

so that ${\displaystyle \lambda *p^{2}+q^{2}}${\displaystyle \lambda *p^{2}+q^{2}}, or its half or quarter, is a factor of N.

Can someone verify whether this is true or not. And whether this math should get into the article.

It seems that finding one ${\displaystyle a^{2}+\lambda *b^{2}=P*Q}${\displaystyle a^{2}+\lambda *b^{2}=P*Q} is easy but finding two with the same lambda is difficult.

The following equation shows this to work:

Solve[
 a^2 + 5 b^2 == 8467 39343 && a > 0 && b > 0, {a, b}, Integers]
{{a -> 16541, b -> 3450}, {a -> 17776, b -> 1851}}
aa = 
 Solve[16541 -ひく 17776 =わ=わ 5 m p && 3450 + 1851 == m q && 
 3450 - 1851 == n p, {m, n, p, q}, Integers]
 {{m -> -19, n -> 123, p -> 13, q -> -279}, {m -> 19, 
 n -> -123, p -> -13, q -> 279}}
now one of the factors is seen, 39343
(1/2) (5 p^2 + q^2) /. aa
{39343, 39343}

This example shows the factoring algorithm works on 3 mod 4 semiprimes, and is thus more general than the more well known Euler factoring algorithm.

James McKee has a paper ^[3] on this type of factorization and claims it is ${\displaystyle \Omega ({\sqrt[{3}]{N}})}${\displaystyle \Omega ({\sqrt[{3}]{N}})}.

Another source that extends Euler's ${\displaystyle x^{2}+D*y^{2}==p*q}${\displaystyle x^{2}+D*y^{2}==p*q} work to ${\displaystyle F*x^{2}+D*y^{2}==p*q}${\displaystyle F*x^{2}+D*y^{2}==p*q} is at url ^[4] . — Preceding unsigned comment added by Endo999 (talk • contribs) 01:55, 2 April 2020 (UTC) [reply ]

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