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)

"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)

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

Another Euler Factorisation method mentioned in Dickson's History of Numbers

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":

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

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

so that $\lambda *p^{2}+q^{2}$, 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 two sets of $a^{2}+\lambda *b^{2}=P*Q$ is easier than finding two sets of $x^{2}+y^{2}=P*Q$.