identité du mois de novembre 2018

Posté le : jeudi 1 novembre 2018 par Vincent Thill
(a^4-2b^4)^4+(2ba^3)^4+[(c+2)ab^3]^4=(a^4+2b^4)^4+[(c+1)ab^3]^4+(cab^3)^4

 

[(c+2)ab^3]^4+(8ba^3)^4+(8a^4-b^4)^4=(cab^3)^4+(8a^4+b^4)^4+[(c+1)ab^3]^4

 

(8a^4+b^4)^4+(a^4-2b^4)^4+(2ba^3)^4=(a^4+2b^4)^4+(8ba^3)^4+(8a^4-b^4)^4

Lorsque    c = 1      ou      c = 7      et  a et b  entiers quelconques

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[a^4d^{(12-3n)}+2b^4d^n]^4+[(c+1)ab^3d^3]^4+[cab^3d^3]^4 =

[a^4d^{(12-3n)}-2b^4d^n]^4+[2ba^3d^{(9-2n)}]^4+[(c+2)ab^3d^3]^4  et

[8a^4d^{(12-3n)}+b^4d^n]^4+[(c+1)ab^3d^3]^4+[cab^3d^3]^4 =

[8a^4d^{(12-3n)}-b^4d^n]^4+[8ba^3d^{(9-2n)}]^4+[(c+2)ab^3d^3]^4

Lorsque       c = 1    ou    c = 7      et   a  et  b  entiers quelconques

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Avec     a, b, c, n et k  entiers quelconques

[a^4b^{(12-3n)}+2c^4b^n]^4+[a^4b^{(12-3k)}-2c^4b^k]^4+[2ca^3b^{(9-2k)}]^4 =

[2ca^3b^{(9-2n)}]^4+[a^4b^{(12-3k)}+2c^4b^k]^4+[a^4b^{(12-3n)}-2c^4b^n]^4

exemple    n = 2   ;   k = 3   ;   a = 5   ;   b = 7    et    c = 11     pgcd = 49

(1529907)^4+(200599)^4+(19250)^4=(209349)^4+(1471343)^4+(943250)^4

[8a^4b^{(12-3n)}+c^4b^n]^4+[8a^4b^{(12-3k)}-c^4b^k]^4+[8ca^3b^{(9-2k)}]^4 =

[8a^4b^{(12-3k)}+c^4b^k]^4+[8a^4b^{(12-3n)}-c^4b^n]^4+[8ca^3b^{(9-2n)}]^4

 

[8a^4b^{(12-3n)}+c^4b^n]^4+[a^4b^{(12-3k)}-2c^4b^k]^4+[2ca^3b^{(9-2k)}]^4 =

[a^4b^{(12-3k)}+2c^4b^k]^4+[8a^4b^{(12-3n)}-c^4b^n]^4+[8ca^3b^{(9-2n)}]^4
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