St. Petersburg Mathematical Journal, 19. cilt,495-765. sayfalarAmerican Mathematical Society, 2008 |
Kitabın içinden
46 sonuçtan 1-3 arası sonuçlar
Sayfa 599
... solutions P ( x ) = ± 1 and Q ( x ) = 0 are ignored . In the Poncelet problem , the polynomial R ( x ) will have ... solution with a polynomial P ( x ) of degree n + 1 if and only if == H ( 3 , n ) = 0 . Therefore , the Cayley criterion ...
... solutions P ( x ) = ± 1 and Q ( x ) = 0 are ignored . In the Poncelet problem , the polynomial R ( x ) will have ... solution with a polynomial P ( x ) of degree n + 1 if and only if == H ( 3 , n ) = 0 . Therefore , the Cayley criterion ...
Sayfa 694
... solutions nonequivalent modulo p3 , and these solutions are already nonequivalent modulo p , but if xq ( p ) = -1 , then the congruence a2 = d ( mod 4p ) is unsolvable , and the congruence ( 4.14 ) has no primitive solutions ( see ...
... solutions nonequivalent modulo p3 , and these solutions are already nonequivalent modulo p , but if xq ( p ) = -1 , then the congruence a2 = d ( mod 4p ) is unsolvable , and the congruence ( 4.14 ) has no primitive solutions ( see ...
Sayfa 942
... solutions are qЄ { 4,9 } . Finally , if t≥ 3 , then only q = 2t with tЄ { 3,4 } can be a solution of the above inequality . Now , we rule out the remaining simple groups for n = 2. We know that the groups C2 ( g ) and B2 ( q ) coincide ...
... solutions are qЄ { 4,9 } . Finally , if t≥ 3 , then only q = 2t with tЄ { 3,4 } can be a solution of the above inequality . Now , we rule out the remaining simple groups for n = 2. We know that the groups C2 ( g ) and B2 ( q ) coincide ...
İçindekiler
A Brudnyi and D Kinzebulatov Uniform subalgebras of L on the unit | 495 |
N A Vavilov Can one see the signs of structure constants? | 519 |
Waldemar Hołubowski A new measure of growth for groups and algebras | 545 |
Telif Hakkı | |
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