Fermat, Goldbach, Pythagoras, Diophantus, perfect numbers |

i^{2} + n^{2} = m^{2}

i^{2} = (m+n)(m-n) ; if p^{2k-1}|(m+n), p≠2, but p^{2k} does not, then p^{2k}|i^{2} => p|(m-n) => p|m,n

Therefor in the simplified case, common factors removed, p^{2k-1}|(m+n), p≠2 => p^{2k}|(m+n).

m = [(x) + (i^{2}/x)]/2 : for x|i^{2} and x,(i^{2}/x) even; or x,i odd

n = [(x) - (i^{2}/x)]/2 : (ibid)

: that is m,n both even or odd, (i^{2}/m+n) even; or m or n odd, not both, i odd

**i: (i n m/p1/p2/...) (...) ...
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[under construction]
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CONSIDER:
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X ^{N} + Y^{N} = Z^{N} whence: (X^{N}/Z-Y) =
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Thence:
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(X ^{N}/Z-Y) - (Y^{N}/Z-X) = (Y-X) (**

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[rechecking-- compose mode]
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(Z-X/Z-Y) > (Y ^{N}/X^{N}) which we normalize over Z, on the z-unit curve:
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(1-x/1-y) > (y ^{N}/x^{N}) where x,y,z == X/Z,Y/Z,(Z/Z=1).
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Minimum value of X: Y=Z-1: X ^{N} = (Z-(Z-1)) *
( Σ_{n+m = N-1} Z^{m} (Z-1)^{n})
&eog; N * (Z-.5)^{N-1}
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Modulus, because X,Y,Z are relative primes:
X ^{N} = Z^{N} mod Y;
Y^{N} = Z^{N} mod X;
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Therefor N even; Z>Y>X :
Z ^{N/2} = ± X^{N/2} mod ... (Y);
Z^{N/2} = ± Y^{N/2} mod ... (X);
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And N odd; Z>Y>X :
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X ^{N}X^{N}
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