关于丢番图方程X^4-Dy^2=1

ON THE DIOPHANTINE EQUATION x^4-Dy^2=1

方程x^4-Dy^2=1国内外许多学者从事了数十年的研究,有人曾宣称再要得出新的结果将是非常困难的.本文对国内的几位主要研究者如著名数学家柯召以及孙琦、曹珍富的一些结果将作出较大改进,去掉某些不必要的、且限制很强的条件,同时简化他们某些定理的证明过程,得出一批新的结果,现发表其中的一部分.

For the Diophantine equationx4-Dy2 = 1 (1)where D>0 and is not a perfect square, well prove the following theorems in this peper:Theoreml. If D=2 pq1…qu,μ≥1, p(?)1(mod 4), q1(?)q2(?)…qμ(?)3 (mod4) (I) there is a α(α≤(?)) ,for which Legendre Symbal(qα/p)=-1,Or(II)P(?)5 (mod 8)D≠D1=210, D≠D2=184030,then (1) has no solutions in positive inleger x, y,if D=D1 or D2,then (1) has (x, y)=(41, 116) or (x, y)=(47321'5219916);(III)2p=a2+b2, a(?)±3 (mod 8) , b(?)±(mod 8),then (1) has no solutions in positive integer x, y.Corollary of theoreml. If D1=2 q1…qμ,μ≥2; D2=2q, qi(?)3 (mod 4) (i=1,…,μ); q≠3 q are primes, then (1) has no Solutions in positive integer x, y. If q=3, D2=6, then (1) has (x, y)=(7, 20).Theonrem 2.(I) The equation (1) with D =pq1…q(?),μ≥1, where p, qi (i=1, 2,…μ) are distinct primes, p(?)1 (mod 4), qi(?)3 (mod 4), then (1) has no solutions in positive integer x, y, if1) there is a α(α≤μ), qα(?)7 (mod 8) , (qα/p)=-1or 2) there is a α and γ (β,γ≤μ) (β≠γ) , r which (qβ/p)=+1 (qγ/p)=-1, qβ=3 (mod 8)(II)If D=pq, p, q are distinct prime, p(?)1 (mod 12) , q(?) 3 (mod 4) ,(q/p)(?)-1,then (1) has no Solutions in positive integer x, y. Corollary 1 of theorem 2. If D=q1…qμ,μ≥1,primes qi(?)3(mod 4) (i=…μ) , then (1) has no solutions in positive integer x, y.Corollary 2 of theorem 2 If D=15D1, D1(?)1(mod 8), qi|D1, qi(?)3 (mod 4) , then (1) has no solutions in positive integer x, y.Theorem 3. (I) The equation (1) with D =pq1…qμ,μ=2k-1, k≥1, has no solutions in positive integer x, y, if 1) p(?)17 (mod 24) , q1…qμ(?)11(mod 24) or2)p(?)5 (mod 24), q1…qμ(?)23 (mod 24).(II)If D=pq, p(?)5 (mod 12),q(?)3 (mod 8) , (p/q)=+1,then(1) has no solutions in positive integer x, y.