Ta có: \(a^5-a=a\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right)=a\left(a-1\right)\left(a+1\right)\left(a^2-4+5\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a^2-4\right)+5a\left(a-1\right)\left(a+1\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)+5a\left(a-1\right)\left(a+1\right)⋮5\)
vì \(a-2;a-1;a;a+1;a+2\) là 5 số nguyên liên tiếp
=> \(a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)⋮5\)
và \(5a\left(a-1\right)\left(a+1\right)⋮5\)
=> \(a^5-a⋮5\)tương tự ta cũng có: \(b^5-b⋮5\) và \(c^5-c⋮5\)
=> \(\left(a^5-a\right)+\left(b^5-b\right)+\left(c^5-c\right)⋮5\)
=> \(\left(a^5+b^5+c^5\right)-\left(a+b+c\right)⋮5\)
=> \(a^5+b^5+c^5⋮5\)