Ta có :
\(a,b\) là số nguyên tố > 3
\(\Leftrightarrow a;b⋮̸\) \(3\)
\(\Leftrightarrow a^2;b^2\) chia \(3\) dư \(1\)
\(\Leftrightarrow a^2-b^2⋮3\)
\(\Leftrightarrowđpcm\)
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
\(\left(a+b\right)\left(a-b\right)=\left(3k+2+3k+2\right)\left(3k+2-3k-2\right)=\left(6k+4\right).0=0⋮3\)
\(\Rightarrow\left(a+b\right)\left(a-b\right)⋮3\Rightarrow a^2-b^2⋮3\rightarrowđpcm\)
