Dễ thấy \(p\) là 1 số nguyên tố lẻ
xét: \(p=3\Leftrightarrow\left\{{}\begin{matrix}p+4=7\\p+8=11\end{matrix}\right.\left(chon\right)\)
Với \(p\) là 1 số nguyên tố \(>3\) thì \(p\) sẽ có dạng \(3t+1\) và \(3t+2\)
Ta có: \(\left\{{}\begin{matrix}p+8=3t+1+8=3t+9=3\left(t+3\right)\\p+4=3t+2+4=3t+6=3\left(t+2\right)\end{matrix}\right.\left(hs\right)\left(l\right)\)
Vậy \(p=3\)
tim n de n^3 - n^2 + n -1 la so nguyen to
cho p; p+20 va p+40 la cac so nguyen to
chung minh rang p+80 cung la so nguyen to
3n+1 va 4n+1 la 2 so nguyen to cung nhau
1.tim tat ca cac so nguyen duong n sao cho tat ca cac so n+1,n+5,n+7,n+13,n+17,n+25,n+37 deu la cac so nguyen to
TRINH BAY DAY DU GIUP MINH NHA
tim so tu nhien nho nhat chi chua thua so nguyen to 2 va 5 biet khi chia no cho 2 thi duoc mot so chinh phuong
Tim cac so nguyen n sao cho:
A=\(\dfrac{n-3}{n+1}\)la so nguyen C=\(\dfrac{2n+3}{n-1}\)la so nguyen
B=\(\dfrac{2n-3}{n+2}\)la so nguyen D=\(\dfrac{-n+5}{n+2}\)la so nguyen
CM
neu p la so nguyen to >3 thi (p-1)(p+1) ⋮ 3 ; ⋮8
tong hieu sau la so nguyen to hay hop so : 5.9.11-2.3.7
Cho p la snt lon hon 3. Biet 8p + 1 cung la snt . Hoi 4p + 1 la so nguyen to hay hop so.
