Hence : a^2 - c^2 = d^2 - b^2

So : (a-c)(a+c) = (d-b)(d+b)

Now, let k = gcd(a-c, d-b)

Then : a-c = k*l and d-b = k*m,

for some l and m, where (l,m) = 1

Then : l(a+c) = m(d+b)

Some icky algebra then shows that

N = [(k/2)^2 + (n/2)^2] * (l^2 + m^2)

where n = (a+c)/m = (b+d)/l

- Choose a number,
`N`, you wish to factor. - Somehow magically express it as the sum of two squares in two different ways.
- Do the algebra shown above.
- Done. (mod details)

- Theoretical runtime:
- O( ? ) - probably slow.

- Time analysis: Not yet available
- Source code: Not yet available

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Back to the pslc home page Paul Herman

pherman@frenchfries.net

Last Updated: May 23, 1997