Gọi d là \(Ư\)CLN( 3n + 1 , 4n + 1 )
\(\Rightarrow3n+1⋮d\)
\(\Rightarrow4n+1⋮d\)
\(\Rightarrow\left(3n+1\right)-\left(4n+1\right)⋮d\)
\(\Rightarrow4\left(3n+1\right)-3\left(4n+1\right)⋮d\)
\(\Rightarrow\left(12n+4\right)-\left(12n+3\right)⋮d\)
\(\Rightarrow1⋮d\)
=> d = 1
Vậy 3n+1 và 4n+1 là hai số nguyên tố cùng nhau
cho p; p+20 va p+40 la cac so nguyen to
chung minh rang p+80 cung la so nguyen to
Cho p la snt lon hon 3. Biet 8p + 1 cung la snt . Hoi 4p + 1 la so nguyen to hay hop so.
Tim so nguyen to p de p+4 va p+8 la so nguyen to
tim n thuoc N de 9n + 24 va 3n +4 nguyen to cung nhau
giai
gia su: UCLN ( 9n + 24;3n+ 4) =d
suy ra 9n +24 chia het cho d va 3n + 4 chia het cho d
suy ra 3(3n+4) chia het cho d
suy ra 9n+12 chia het cho d
suy ra ( 9n+24) - ( 9n + 12) chia het cho d
12 chia het cho d suy ra d=1;d=2; d=3; d=4; d=6; d= 12
( tim cach loai di d=2; d=3 ; d=4; d= 6; d=12)
chung minh rang hieu cua 2 phan so toi gian la 1 so nguyen thi 2 phan so co mau bang nhau hoac doi nhau
tim n de n^3 - n^2 + n -1 la so nguyen to
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
cho B la so tu nhien . biet B = ax -by ; a;b la cac so nguyen to khac nhau ;x,y la hai so tu nhien khac 0 . B2 co 15 uoc hoi B3 co bao nhieu uoc
