PQT
Cho $p$ nguyên tố, gọi $S_1(p)= \{ (a,b,c) \in \mathbb{Z}^3, p|a^2b^2+b^2c^2+c^2a^2+1 \}$ và $S_2(p)= \{ (a,b,c) \in \mathbb{Z}^3, p|a^2b^2c^2(a^2+b^2+c^2+a^2b^2c^2) \}$. Tìm tất cả các số nguyên tố $p$ thoả mãn $S_1(p) \subset S_2(p)$.
Bài viết đã được chỉnh sửa nội dung bởi Jinbe: 24-10-2013 - 12:19
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).
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cái này là MR mà
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