Giải phương trình nghiệm nguyên $$y(x + y) = x^3- 7x^2 + 11x - 3.$$
Giải phương trình nghiệm nguyên $$y(x + y) = x^3- 7x^2 + 11x - 3.$$
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”).
Bài toán thành: $(2y+x)^{2}=(x-2)(4x^{2}-19x+6)$ . Gọi $d = GCD(x-2;4x^{2}-19x+6) \epsilon N^{*}$ . Chứng minh được $16\vdots d\rightarrow d\epsilon U(16)=\begin{Bmatrix} 1,2,4,8,16 \end{Bmatrix}$ . Nếu $d\epsilon \begin{Bmatrix} 1,4,16 \end{Bmatrix} \rightarrow \left\{\begin{matrix}x-2=m^{2} \\ 4x^{2}-19x+6=n^{2} (4x^{2}-19x+6> 0\forall x) \end{matrix}\right.\Leftrightarrow 265=(8x-19-4n)(8x-19+4n)\rightarrow x=6,y\epsilon \begin{Bmatrix} 3,-9 \end{Bmatrix}$ đến đây giải ra. Nếu $d\epsilon \begin{Bmatrix}2,8 \end{Bmatrix}\rightarrow \left\{\begin{matrix}x-2=2m^{2} \\ 4x^{2}-19x+6=2n^{2} \end{matrix}\right.$ Đặt $x=2k\epsilon Z \Leftrightarrow \left\{\begin{matrix} k=m^{2}+1 \\ 8k^{2}-19k+3=n^{2} \end{matrix}\right. \rightarrow k-1\equiv 0,1,4(mod5)\rightarrow k\equiv 0,2,1(mod5)$ .
$k\equiv0(mod5)\rightarrow n^{2}\equiv 3(mod5);k\equiv 1(mod5)\rightarrow n^{2} \equiv2(mod5);k\equiv 2(mod5)\rightarrow n^{2} \equiv 3(mod5).$ (loại ). Bài này hay nhưng không quá khó
Bài viết đã được chỉnh sửa nội dung bởi Nguyen Bao Khanh: 08-04-2023 - 21:23
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