Gọi $\|1,4,4,5;n\|$ là số nghiệm nguyên không âm của phương trình $x_1+4(x_2+x_3)+5x_4=n$
Chứng minh rằng:
\begin{equation}\label{e1}
\|1,4,4,5;n\|=\fl{\dfrac{2n^3+42n^2+265n+30(n+3)[(n+1\!\!\mod 4)-(n\!\!\mod 4)]-15n(-1)^n+960)}{960}}
\end{equation}
\begin{equation}\label{e2}
\|1,4,4,5;n\|=\dfrac{n^3+21n^2+140n+480}{480},\quad (n\equiv 0,4\!\!\!\pmod{20})
\end{equation}
\begin{equation}\label{e3}
\|1,4,4,5;n\|=\dfrac{n^3+21n^2+140n+384}{480},\quad (n\equiv 8\!\!\!\pmod{20})
\end{equation}
\begin{equation}\label{e4}
\|1,4,4,5;n\|=\dfrac{n^3+21n^2+140n+288}{480},\quad (n\equiv 12,16\!\!\!\pmod{20})
\end{equation}
\begin{equation}\label{e5}
\|1,4,4,5;n\|=\dfrac{n^3+21n^2+155n+303}{480},\quad (n\equiv 1,17\!\!\!\pmod{20})
\end{equation}
\begin{equation}\label{e6}
\|1,4,4,5;n\|=\dfrac{n^3+21n^2+155n+495}{480},\quad (n\equiv 5,9\!\!\!\pmod{20})
\end{equation}
\begin{equation}\label{e7}
\|1,4,4,5;n\|=\dfrac{n^3+21n^2+155n+399}{480},\quad (n\equiv 13\!\!\!\pmod{20})
\end{equation}
\begin{equation}\label{e8}
\|1,4,4,5;n\|=\dfrac{n^3+21n^2+140n+108}{480},\quad (n\equiv 2,6\!\!\!\pmod{20})
\end{equation}
\begin{equation}\label{e9}
\|1,4,4,5;n\|=\dfrac{n^3+21n^2+140n+300}{480},\quad (n\equiv 10,14\!\!\!\pmod{20})
\end{equation}
\begin{equation}\label{e10}
\|1,4,4,5;n\|=\dfrac{n^3+21n^2+140n+204}{480},\quad (n\equiv 18\!\!\!\pmod{20})
\end{equation}
\begin{equation}\label{e11}
\|1,4,4,5;n\|=\dfrac{n^3+21n^2+95n-21}{480},\quad (n\equiv 3\!\!\!\pmod{20})
\end{equation}
\begin{equation}\label{e12}
\|1,4,4,5;n\|=\dfrac{n^3+21n^2+95n-117}{480},\quad (n\equiv 7,11\!\!\!\pmod{20})
\end{equation}
\begin{equation}\label{e13}
\|1,4,4,5;n\|=\dfrac{n^3+21n^2+95n+75}{480},\quad (n\equiv 15,19\!\!\!\pmod{20})
\end{equation}
Bạn có thể thử sức với một trong số các công thức trên.
Bài viết đã được chỉnh sửa nội dung bởi hxthanh: 26-03-2024 - 07:25
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