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When was F. De Samara to A. G. A. created?

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Anonymous

 ∙ 11y ago
Updated: 8/21/2019

F. De Samara to A. G. A. was created in 1838.

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 ∙ 11y ago

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Who can solve FLT short?

Le dernier théorème de Pierre de Fermat . (x, y, z, n) l'ensemble de ( N+ )^4. n> 2. ( a ) l'ensemble de Z F est la fonction de (a). F (a) = [a (a +1) / 2] ^ 2 F (0) = 0 et F (-1) = 0. Considérons deux équations. F (z) = F (x) + F (y) F (z-1) = F (x-1) + F (y-1) Nous avons une inférence chaîne F (z) = F (x) + F (y) équivalent F (z-1) = F (x-1) + F (y-1) F (z) = F (x) + F (y) en déduire F (z-1) = F (x-1) + F (y-1) F (z-x-1) = F (x-x-1) + F (y-x-1) en déduire F (z-x-2) = F (x-x-2) + F (y-x-2) nous voyons F (z-x-1) = F (x-x-1) + F (y-x-1) F (z-x-1) = F (-1) + F (y-x-1) F (z-x-1) = 0 + F (y-x-1) donner z = y et F (z-x-2) = F (x-x-2) + F (y-x-2) F (z-x-2) = F (-2) + F (y-x-2) F (z-x-2) = 1 + F (y-x-2) donner z = / = y. de sorte F (z-x-1) = F (x-x-1) + F (y-x-1) ne pas en déduire F (z-x-2) = F (x-x-2) + F (y-x-2) de sorte F (z) = F (x) + F (y) ne pas en déduire F (z-1) = F (x-1) + F (y-1) de sorte F (z) = F (x) + F (y) n'est pas équivalente F (z-1) = F (x-1) + F (y-1) Donc avoir deux cas. [F (x) + F (y)] = F (z) et F (x-1) + F (y-1)] = / = F (z-1) ou vice versa de sorte [F (x) + F (y)] - [F (x-1) + F (y-1)] = / = F (z)-F (z-1). Ou F (x)-F (x-1) + F (y)-F (y-1) = / = F (z)-F (z-1). nous voyons F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. de sorte x 3 + y ^3 =/= z ^ 3. n> 2. . Similaire. Nous avons une inférence chaîne G (z) * F (z) = G (x) * F (x) + G (y) * F (y) équivalente G (z) * F (z-1) = G (x) * F (x -1) + G (y) * F (y-1) G (z) * F (z) = G (x) * F (x) + G (y) * F (y) en déduire G (z) * F (z-1) = G (x) * F (x -1) + G (y) * F (y-1) G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y-x-1) * F (y) en déduire G (z) * F (z-x-2) = G ( x) * F (x-x-2) + G (y) * F (y-x-2) nous voyons G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y) * F (y-x-1) G (z) * F (z-x-1) = G (x) * F (-1) + G (y) * F (y-x-1) G (z) * F (z-x-1) = 0 + G (y) * F (y-x-1) donner z = y. et G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2) G (z) * F (z-x-2) = G (x) * F (-2) + G (y) * F (y-x-2) G (z) * F (z-x-2) = G (x) + G (y) * F (y-x-2) x> 0 en déduire G (x)> 0. donner z = / = y. de sorte G (z) * F (zx-1) = G (x) * F (xx-1) + G (yx-1) * F (y) ne pas en déduire G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2) de sorte G (z) * F (z) = G (x) * F (x) + G (y) * F (y) ne pas en déduire G (z) * F (z-1) = G (x) * F (x-1) + G (y) * F (y-1) de sorte G (z) * F (z) = G (x) * F (x) + G (y) * F (y) n'est pas équivalente G (z) * F (z-1) = G (x) * F (x-1) + G (y) * F (y-1) Donc avoir deux cas [G (x) * F (x) + G (y) * F (y)] = G (z) * F (z) et [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z-1) * F (z-1) ou vice versa. de sorte [G (x) * F (x) + G (y) * F (y)] - [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z) * [F (z)-F (z-1)]. Ou G (x) * [F (x) - F (x-1)] + G (y) * [F (y)-F (y-1)] = / = G (z) * [F (z) -F (z-1).] nous voyons x ^ n = G (x) * [F (x)-F (x-1)] y ^ n = G (y) * [F (y)-F (y-1)] z ^ n = G (z) * [F (z)-F (z-1)] de sorte x ^ n + y ^ n = / = z ^ n Le bonheur et la paix Tran Tan Cuong

Who is person who solve shortest Fermat?

To: trantancuong21@yahoo.com Le dernier théorème de Pierre de Fermat . (x, y, z, n) l'ensemble de ( N+ )^4. n> 2. ( a ) l'ensemble de Z F est la fonction de (a). F (a) = [a (a +1) / 2] ^ 2 F (0) = 0 et F (-1) = 0. Considérons deux équations. F (z) = F (x) + F (y) F (z-1) = F (x-1) + F (y-1) Nous avons une inférence chaîne F (z) = F (x) + F (y) équivalent F (z-1) = F (x-1) + F (y-1) F (z) = F (x) + F (y) en déduire F (z-1) = F (x-1) + F (y-1) F (z-x-1) = F (x-x-1) + F (y-x-1) en déduire F (z-x-2) = F (x-x-2) + F (y-x-2) nous voyons F (z-x-1) = F (x-x-1) + F (y-x-1) F (z-x-1) = F (-1) + F (y-x-1) F (z-x-1) = 0 + F (y-x-1) donner z = y et F (z-x-2) = F (x-x-2) + F (y-x-2) F (z-x-2) = F (-2) + F (y-x-2) F (z-x-2) = 1 + F (y-x-2) donner z = / = y. de sorte F (z-x-1) = F (x-x-1) + F (y-x-1) ne pas en déduire F (z-x-2) = F (x-x-2) + F (y-x-2) de sorte F (z) = F (x) + F (y) ne pas en déduire F (z-1) = F (x-1) + F (y-1) de sorte F (z) = F (x) + F (y) n'est pas équivalente F (z-1) = F (x-1) + F (y-1) Donc avoir deux cas. [F (x) + F (y)] = F (z) et F (x-1) + F (y-1)] = / = F (z-1) ou vice versa de sorte [F (x) + F (y)] - [F (x-1) + F (y-1)] = / = F (z)-F (z-1). Ou F (x)-F (x-1) + F (y)-F (y-1) = / = F (z)-F (z-1). nous voyons F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. de sorte x 3 + y ^3 =/= z ^ 3. n> 2. . Similaire. Nous avons une inférence chaîne G (z) * F (z) = G (x) * F (x) + G (y) * F (y) équivalente G (z) * F (z-1) = G (x) * F (x -1) + G (y) * F (y-1) G (z) * F (z) = G (x) * F (x) + G (y) * F (y) en déduire G (z) * F (z-1) = G (x) * F (x -1) + G (y) * F (y-1) G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y-x-1) * F (y) en déduire G (z) * F (z-x-2) = G ( x) * F (x-x-2) + G (y) * F (y-x-2) nous voyons G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y) * F (y-x-1) G (z) * F (z-x-1) = G (x) * F (-1) + G (y) * F (y-x-1) G (z) * F (z-x-1) = 0 + G (y) * F (y-x-1) donner z = y. et G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2) G (z) * F (z-x-2) = G (x) * F (-2) + G (y) * F (y-x-2) G (z) * F (z-x-2) = G (x) + G (y) * F (y-x-2) x> 0 en déduire G (x)> 0. donner z = / = y. de sorte G (z) * F (zx-1) = G (x) * F (xx-1) + G (yx-1) * F (y) ne pas en déduire G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2) de sorte G (z) * F (z) = G (x) * F (x) + G (y) * F (y) ne pas en déduire G (z) * F (z-1) = G (x) * F (x-1) + G (y) * F (y-1) de sorte G (z) * F (z) = G (x) * F (x) + G (y) * F (y) n'est pas équivalente G (z) * F (z-1) = G (x) * F (x-1) + G (y) * F (y-1) Donc avoir deux cas [G (x) * F (x) + G (y) * F (y)] = G (z) * F (z) et [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z-1) * F (z-1) ou vice versa. de sorte [G (x) * F (x) + G (y) * F (y)] - [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z) * [F (z)-F (z-1)]. Ou G (x) * [F (x) - F (x-1)] + G (y) * [F (y)-F (y-1)] = / = G (z) * [F (z) -F (z-1).] nous voyons x ^ n = G (x) * [F (x)-F (x-1)] y ^ n = G (y) * [F (y)-F (y-1)] z ^ n = G (z) * [F (z)-F (z-1)] de sorte x ^ n + y ^ n = / = z ^ n Le bonheur et la paix Tran Tan Cuong

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a-g-f-f-f-f-f-f-f-g-a-g-a-g-f-f-f-f-f-f-f-f-f-a-g-g-a-g-f-f-g-a-a-a-a-a-a-g-f

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a-g-f-f-f-f-f-f-f-g-a-g-a-g-f-f-f-f-f-f-f-f-f-a-g-g-a-g-f-f-g-a-a-a-a-a-a-g-f that is only chorus

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Baby by: Justin BieberF G F F A G F E D E D F A G F E D E D F A G F G G F EF C2 A G A F C2 A G F C2 A G A F C2 A G F C2 C2 A G F C2 C2 A GF F A G A G A G A G A G F C2 A G A C2 A G F C2 G A F F G F F F A A G F GF F G G G G G G A G F F G FChorus: A G A G A G C2 G A G A G A G D2 G A G A G A G C2 A G A A A G FA G A G A G C2 G A G A G A G D2 G A G A G A G C2 A G A A A G F-Rossele-Send more requests @ycel_gandah@Yahoo.comTy!

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