In this paper we prove in a new way, the well known result, that Fermat’s equation a<sup>4</sup> + b<sup>4</sup> = c<sup>4</sup>, is not solvable in ℕ , when abc≠0 . To show this ...In this paper we prove in a new way, the well known result, that Fermat’s equation a<sup>4</sup> + b<sup>4</sup> = c<sup>4</sup>, is not solvable in ℕ , when abc≠0 . To show this result, it suffices to prove that: ( F 0 ): a 1 4 + ( 2 s b 1 ) 4 = c 1 4 , is not solvable in ℕ , (where a 1 , b 1 , c 1 ∈2ℕ+1 , pairwise primes, with necessarly 2≤s∈ℕ ). The key idea of our proof is to show that if (F<sub>0</sub>) holds, then there exist α 2 , β 2 , γ 2 ∈2ℕ+1 , such that ( F 1 ): α 2 4 + ( 2 s−1 β 2 ) 4 = γ 2 4 , holds too. From where, one conclude that it is not possible, because if we choose the quantity 2 ≤ s, as minimal in value among all the solutions of ( F 0 ) , then ( α 2 ,2 s−1 β 2 , γ 2 ) is also a solution of Fermat’s type, but with 2≤s−1<s , witch is absurd. To reach such a result, we suppose first that (F<sub>0</sub>) is solvable in ( a 1 ,2 s b 1 , c 1 ) , s ≥ 2 like above;afterwards, proceeding with “Pythagorician divisors”, we creat the notions of “Fermat’s b-absolute divisors”: ( d b , d ′ b ) which it uses hereafter. Then to conclude our proof, we establish the following main theorem: there is an equivalence between (i) and (ii): (i) (F<sub>0</sub>): a 1 4 + ( 2 s b 1 ) 4 = c 1 4 , is solvable in ℕ , with 2≤s∈ℕ , ( a 1 , b 1 , c 1 )∈ ( 2ℕ+1 ) 3 , coprime in pairs. (ii) ∃( a 1 , b 1 , c 1 )∈ ( 2ℕ+1 ) 3 , coprime in pairs, for wich: ∃( b ′ 2 , b 2 , b ″ 2 )∈ ( 2ℕ+1 ) 3 coprime in pairs, and 2≤s∈ℕ , checking b 1 = b ′ 2 b 2 b ″ 2 , and such that for notations: S=s−λ( s−1 ) , with λ∈{ 0,1 } defined by c 1 − a 1 2 ≡λ( mod2 ) , d b =gcd( 2 s b 1 , c 1 − a 1 )= 2 S b 2 and d ′ b = 2 s−S b ′ 2 = 2 s B 2 d b , where ( 2 s B 2 ) 2 =gcd( b 1 2 , c 1 2 − a 1 2 ) , the following system is checked: { c 1 − a 1 = d b 4 2 2+λ = 2 2−λ ( 2 S−1 b 2 ) 4 c 1 + a 1 = 2 1+λ d ′ b 4 = 2 1+λ ( 2 s−S b ′ 2 ) 4 c 1 2 + a 1 2 =2 b ″ 2 4;and this system implies: ( b 1−λ,2 4 ) 2 + ( 2 4s−3 b λ,2 4 ) 2 = ( b ″ 2 2 ) 2;where: ( b 1−λ,2 , b λ,2 , b ″ 2 )={ ( b ′ 2 , b 2 , b ″ 2 ) if λ=0 ( b 2 , b ′ 2 , b ″ 2 ) if λ=1;From where, it is quite easy to conclude, following the method explained above, and which thus closes, part I, of this article. .展开更多
文摘In this paper we prove in a new way, the well known result, that Fermat’s equation a<sup>4</sup> + b<sup>4</sup> = c<sup>4</sup>, is not solvable in ℕ , when abc≠0 . To show this result, it suffices to prove that: ( F 0 ): a 1 4 + ( 2 s b 1 ) 4 = c 1 4 , is not solvable in ℕ , (where a 1 , b 1 , c 1 ∈2ℕ+1 , pairwise primes, with necessarly 2≤s∈ℕ ). The key idea of our proof is to show that if (F<sub>0</sub>) holds, then there exist α 2 , β 2 , γ 2 ∈2ℕ+1 , such that ( F 1 ): α 2 4 + ( 2 s−1 β 2 ) 4 = γ 2 4 , holds too. From where, one conclude that it is not possible, because if we choose the quantity 2 ≤ s, as minimal in value among all the solutions of ( F 0 ) , then ( α 2 ,2 s−1 β 2 , γ 2 ) is also a solution of Fermat’s type, but with 2≤s−1<s , witch is absurd. To reach such a result, we suppose first that (F<sub>0</sub>) is solvable in ( a 1 ,2 s b 1 , c 1 ) , s ≥ 2 like above;afterwards, proceeding with “Pythagorician divisors”, we creat the notions of “Fermat’s b-absolute divisors”: ( d b , d ′ b ) which it uses hereafter. Then to conclude our proof, we establish the following main theorem: there is an equivalence between (i) and (ii): (i) (F<sub>0</sub>): a 1 4 + ( 2 s b 1 ) 4 = c 1 4 , is solvable in ℕ , with 2≤s∈ℕ , ( a 1 , b 1 , c 1 )∈ ( 2ℕ+1 ) 3 , coprime in pairs. (ii) ∃( a 1 , b 1 , c 1 )∈ ( 2ℕ+1 ) 3 , coprime in pairs, for wich: ∃( b ′ 2 , b 2 , b ″ 2 )∈ ( 2ℕ+1 ) 3 coprime in pairs, and 2≤s∈ℕ , checking b 1 = b ′ 2 b 2 b ″ 2 , and such that for notations: S=s−λ( s−1 ) , with λ∈{ 0,1 } defined by c 1 − a 1 2 ≡λ( mod2 ) , d b =gcd( 2 s b 1 , c 1 − a 1 )= 2 S b 2 and d ′ b = 2 s−S b ′ 2 = 2 s B 2 d b , where ( 2 s B 2 ) 2 =gcd( b 1 2 , c 1 2 − a 1 2 ) , the following system is checked: { c 1 − a 1 = d b 4 2 2+λ = 2 2−λ ( 2 S−1 b 2 ) 4 c 1 + a 1 = 2 1+λ d ′ b 4 = 2 1+λ ( 2 s−S b ′ 2 ) 4 c 1 2 + a 1 2 =2 b ″ 2 4;and this system implies: ( b 1−λ,2 4 ) 2 + ( 2 4s−3 b λ,2 4 ) 2 = ( b ″ 2 2 ) 2;where: ( b 1−λ,2 , b λ,2 , b ″ 2 )={ ( b ′ 2 , b 2 , b ″ 2 ) if λ=0 ( b 2 , b ′ 2 , b ″ 2 ) if λ=1;From where, it is quite easy to conclude, following the method explained above, and which thus closes, part I, of this article. .
