Để P là số nguyên tố thì n+ 4 \(⋮\)2n-1
\(\frac{n+4}{2n-1}\)= \(\frac{2\left(n+4\right)}{2n-1}\)= \(\frac{2n+8}{2n-1}\)= \(\frac{2n-1+9}{2n-1}\)= \(\frac{9}{2n-1}\)=> 9 \(⋮\)2n-1
=> 2n-1 \(\in\)Ư(9)= { 1;3 ; 9; -1; -3; -9}
=> 2n \(\in\){ 2; 4; 10; 0; -2; -8}
=> n \(\in\){ 1;2;5; 0; -1; -4}
Vậy...
\(P=\frac{n+4}{2n-1}\)
\(\Leftrightarrow n+4⋮2n-1\)
\(\Leftrightarrow2\left(n+4\right)⋮2n-1\)
\(\Leftrightarrow2n+8⋮2n-1\)
\(\Leftrightarrow2n-1+9⋮2n-1\)
Vì \(2n-1⋮2n-1\)
\(\Leftrightarrow9⋮2n-1\)
\(\Leftrightarrow2n-1\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
Ta lập bảng xét giá trị
| 2n-1 | 1 | -1 | 3 | -3 | 9 | -9 |
| 2n | 2 | 0 | 4 | -2 | 10 | -8 |
| n | 1 | 0 | 2 | -1 | 5 | -4 |
