[ASK]Show that the sum of the fifth powers of these numbers is divisible by 5

Monoxdifly

Well-known member
The sum of ten integers is 0. Show that the sum of the fifth powers of these numbers is divisible by 5.

For this one I don't know what I have to do at all other than brute-forcing which may even be impossible.

castor28

Well-known member
MHB Math Scholar
The sum of ten integers is 0. Show that the sum of the fifth powers of these numbers is divisible by 5.

For this one I don't know what I have to do at all other than brute-forcing which may even be impossible.
Hi Monoxdifly ,

The number of integers does not matter. This would be a great time to use Fermat's Little Theorem : for any integer $a$, $a^5\equiv a\pmod5$, since $5$ is prime.

Monoxdifly

Well-known member
Hi Monoxdifly ,

The number of integers does not matter. This would be a great time to use Fermat's Little Theorem : for any integer $a$, $a^5\equiv a\pmod5$, since $5$ is prime.
So you mean that it applies to any number? That "The sum of ten integers is 0" is just a distraction, then?

castor28

Well-known member
MHB Math Scholar
So you mean that it applies to any number? That "The sum of ten integers is 0" is just a distraction, then?
Hi Monoxdifly ,

The distraction is the fact that there are ten integers. The fact that the sum is $0$ is essential.

More precisely, if the integers are $a, b, c, \ldots$, Fermat's theorem tells us that $a^5\equiv a, b^5\equiv b\ldots\pmod5$.

We have therefore $a^5 + b^5 + \cdots\equiv a + b + \cdots\pmod 5$. As we are told that $a + b +\cdots=0\equiv0\pmod5$, we conclude that $a^5 + b^5 + \cdots\equiv0\pmod5$, which means that the sum is divisible by $5$.

Monoxdifly

Well-known member
I think I kinda get a grasp here. Thanks for your help.