On 8/24/2011 11:57 AM, Bruno Marchal wrote:

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Nu = ((ZUY)^2 + U)^2 + Y ELG^2 + Al = (B - XY)Q^2 Qu = B^(5^60) La + Qu^4 = 1 + LaB^5 Th + 2Z = B^5 L = U + TTh E = Y + MTh N = Q^16R = [G + EQ^3 + LQ^5 + (2(E - ZLa)(1 + XB^5 + G)^4 + LaB^5 + +LaB^5Q^4)Q^4](N^2 -N)+ [Q^3 -BL + L + ThLaQ^3 + (B^5 - 2)Q^5] (N^2 - 1) P = 2W(S^2)(R^2)N^2 (P^2)K^2 - K^2 + 1 = Ta^2 4(c - KSN^2)^2 + Et = K^2 K = R + 1 + HP - H A = (WN^2 + 1)RSN^2 C = 2R + 1 Ph D = BW + CA -2C + 4AGa -5Ga D^2 = (A^2 - 1)C^2 + 1 F^2 = (A^2 - 1)(I^2)C^4 + 1 (D + OF)^2 = ((A + F^2(D^2 - A^2))^2 - 1)(2R + 1 + JC)^2 + 1Thanks to Jones, Matiyasevitch. Some number Nu verifying that systemof diophantine equations (the variables are integers) are "LĂ¶bianstories", on which the machine's first person indeterminacy will bedistributed.We don't even need to go farer than the polynomial equations todescribe the ROE.

`I'm reminded of the apocryphal story of Euler being asked by Catherine`

`the Great to counter Diederot who was trying to convert the Russian`

`court to atheism. Euler wrote "e^(i*pi) + 1 = 0 therefore God exists."`

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