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

Monoxdifly

Well-known member
Aug 6, 2015
271
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
Oct 18, 2017
245
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
Aug 6, 2015
271

castor28

Well-known member
MHB Math Scholar
Oct 18, 2017
245
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
Aug 6, 2015
271
I think I kinda get a grasp here. Thanks for your help.