Xet \(p>3\Rightarrow\orbr{\begin{cases}p=3k+1\\p=3k+2\end{cases}}\)
Xet TH \(p=3k+1=>p+14=3k+15=3\left(k+5\right)\)
=> p khong nguyen to
Xet TH \(p=3k+2\Rightarrow p+10=3k+12=3\left(k+4\right)\)
=> p khong nguyen to
Neu \(p< 3=>\hept{\begin{cases}p=0\\p=1\\p=2\end{cases}}\) thay vao p+10 va p+14 dau ko thoa man
Neu p=3 thay vao p+10 va p+14 ta thay thoa man
Vay p =3