Để \(\left(n+3\right)\left(n+1\right)\in P\Rightarrow\orbr{\begin{cases}n+3=1\\n+1=1\end{cases}}\)
Mà \(n+1< n+3\Rightarrow n+1=1\Rightarrow n=0\)
Vậy ...
tim so tu nhien n de (n+3)(n+1) la so nguyen to
tim tat ca cac so tu nhien n de cac so sau la so nguyen to :n+1,n+3,n+7,n+9,n+13,n+15
tim so tu nhien n de 4n+n^2 la so nguyen to
tim so tu nhien n de 2^3n-2 la so nguyen to
a] tim so tu nhien k de 3.k la so nguyen to
b tim so tu nhien k de 7 .k la so nguyen to
a) Tim so tu nhien k de 3.k la so nguyen to
b) Tim so tu nhien k de 7.k la so nguyen to
tim so tu nhien k de 3.k la so nguyen to
tim so tu nhien 7.k la so nguyen to
a) tim so tu nhien k de 3 . k la so nguyen to
b) tim so tu nhien k de 7 . k la so nguyen to
tim so tu nhien n de 2n+1 va 7n+2 la hai so nguyen to cung nhau
