Both a and $a^5$ are even or odd so $a^5-a$
Now if a is multiple of 5 $a^5$ is multiple of 5. so $a^5-a$ is divisible by 5
If a is not multiple of 5 then as per FLT $a^5 \equiv a \pmod 5$ so $a^5-a$ is divisible by 5
In both cases $a^5−a$ is divisible by 10
