More Pythagoras and factoring

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• Just playing with this further there is more note worthy points. Given a number n which is the product of two factors both of which are odd you can derive at
Message 1 of 1 , Dec 2, 2001
Just playing with this further there is more note worthy points.
Given a number n which is the product of two factors both of which are odd
you can derive at least 4 pythagoras triangles with each length being whole.
One of these we've already discussed, one of them is of no major
important(**) but the other two are:

21 -> 28 -> 35
21 -> 72 -> 75

15 -> 20 -> 25
15 -> 36 -> 39

33 -> 44 -> 55
33 -> 180 -> 183

35 -> 84 -> 91
35 -> 120 -> 125

These dimensions are for such p. triangles but not that when you subtract
the two dervied lengths from given number they produce the factors
e.g.
77 -> 264 -> 275
77 -> 420 -> 427

275 - 264 = 11
427 - 420 = 7

7 * 11 = 77

** the other right angled triangle that can be dervied obtains its side
lengths from (n^2) /2 + 0.5 and (n^2) /2 - 0.5
e.g.
85 -> 3612 -> 3613

85 ^ 2 = 7225
7225 /2 = 3612.5

In terms of "striking it lucky" this helps as we now not just have one
target to hit but three - all of which can be used to get the factors out.

Cheers,
David Litchfield
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