pythagodiophantine numbers

Fermat, Goldbach, Pythagoras, Diophantus, perfect numbers

i2 + n2 = m2

i2 = (m+n)(m-n) ; if p2k-1|(m+n), p≠2, but p2k does not, then p2k|i2 => p|(m-n) => p|m,n
Therefor in the simplified case, common factors removed, p2k-1|(m+n), p≠2 => p2k|(m+n).

m = [(x) + (i2/x)]/2 : for x|i2 and x,(i2/x) even; or x,i odd
n = [(x) - (i2/x)]/2 : (ibid)
: that is m,n both even or odd, (i2/m+n) even; or m or n odd, not both, i odd

i: (i n m/p1/p2/...) (...) ...

[under construction]

CONSIDER:

XN + YN = ZN whence: (XN/Z-Y) = Σn+m = N-1 Zm Yn which specifies some factor(s) of XN, reducing it Diophantine-style.

Thence:

(XN/Z-Y) - (YN/Z-X) = (Y-X) (Σi+n+m = N-2 XiYnZm) which relates the sizes of solutions (XN/Z-Y) > (YN/Z-X) to the tendency of solutions of XN + YN = ZN: while XN is the smaller term, yet (XN/Z-Y) must be the larger, by much larger than (Y-X) (N(N-1)/2) XN-2 as both Y,Z are larger than X itself-- which makes this comparative sizably similar to the original equation to be solved;-- however, (Y-X) is not a new factor, but does relate it to higher order (N+1) differences:

[rechecking-- compose mode]

(Z-X/Z-Y) > (YN/XN) which we normalize over Z, on the z-unit curve:

(1-x/1-y) > (yN/xN) where x,y,z == X/Z,Y/Z,(Z/Z=1).

Minimum value of X: Y=Z-1: XN = (Z-(Z-1)) * (Σn+m = N-1 Zm (Z-1)n)
&eog; N * (Z-.5)N-1

Modulus, because X,Y,Z are relative primes:
XN = ZN mod Y;
YN = ZN mod X;

Therefor N even; Z>Y>X :
ZN/2 = ± XN/2 mod ... (Y);
ZN/2 = ± YN/2 mod ... (X);

And N odd; Z>Y>X :

XNXN

Grand-Admiral Petry
'Majestic Service in a Solar System'
Nuclear Emergency Management

2004 GrandAdmiralPetry@Lanthus.net