Question

[Nguyễn Thanh Hữu]

Cho a, b nguyên tố lớn hơn 3

Chứng tỏ a² - b² chia hết cho 3

Answer 1

[Nguyễn Thanh Hằng]

Ta có :

\(a,b\) là số nguyên tố > 3

\(\Leftrightarrow a;b⋮̸\) \(3\)

\(\Leftrightarrow a^2;b^2\) chia \(3\) dư \(1\)

\(\Leftrightarrow a^2-b^2⋮3\)

\(\Leftrightarrowđpcm\)

Answer 2

[Mashiro Shiina]

Ta có: Nếu a;b là các số nguyên tố lớn hơn 3 thì sẽ có dạng 3k+1 ;3k+2

Dựa vào HĐT số 3 ta có:
\(a^2-b^2=\left(a+b\right)\left(a-b\right)\)

Nếu:

a=3k+1;b=3k+2

\(\left(a+b\right)\left(a-b\right)=\left(3k+1+3k+2\right)\left(3k+1-3k+2\right)=\left(6k+3\right).-1=-\left(6k+3\right)⋮3\)a=3k+2;b=3k+1

\(\left(a+b\right)\left(a-b\right)=\left(3k+2+3k+1\right)\left(3k+2-3k-1\right)=\left(6k+3\right).1⋮3\)a=3k+1;b=3k+1

\(\left(a+b\right)\left(a-b\right)=\left(3k+1+3k+1\right)\left(3k+1-3k-1\right)=\left(6k+2\right).0=0⋮3\)a=3k+2;b=3k+2

\(\left(a+b\right)\left(a-b\right)=\left(3k+2+3k+2\right)\left(3k+2-3k-2\right)=\left(6k+4\right).0=0⋮3\)

\(\Rightarrow\left(a+b\right)\left(a-b\right)⋮3\Rightarrow a^2-b^2⋮3\rightarrowđpcm\)