Ta thấy : \(a^5-a=a\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right).\)
\(=a\left(a-1\right)\left(a+1\right)\left(a^2-4+5\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a^2-4\right)+5a\left(a-1\right)\left(a+1\right)\)
\(=\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)+5a\left(a-1\right)\left(a+1\right)\)
Ta có :\(\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)\)là tích 5 số tự nhiên liên tiếp :
\(\Rightarrow\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)\)\(⋮\)\(5\)và cũng \(⋮\)\(6\)( cũng là 3 số tự nhiên liên tiếp )
\(\Rightarrow\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)\)\(⋮\)\(30\)\(\left(1\right)\)
Ta lại có : \(5\)\(⋮\)\(5\)và \(\left(a-1\right)a\left(a+1\right)\)\(⋮\)\(6\)
\(\Rightarrow5a\left(a-1\right)\left(a+1\right)\)\(⋮\)\(30\)\(\left(2\right)\)
Từ ( 1 ) và ( 2 ) \(\Rightarrow\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)+5a\left(a-1\right)\left(a+1\right)\)\(⋮\)\(30\)
Hay \(a^5-a\)\(⋮\)\(30\)
Tương tự \(b^5-b\)và \(c^5-c\)cũng chia hết cho 30
\(\Rightarrow a^5+b^5+c^5-\left(a+b+c\right)\)\(⋮\)\(30\)
Mà \(a+b+c\)\(⋮\)\(30\)
\(\Rightarrow a^5+b^5+c^5\)\(⋮\)\(30\)\(\left(đpcm\right)\)
