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

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The discussion focuses on proving that the sum of the fifth powers of ten integers, which sum to zero, is divisible by 5. Fermat's Little Theorem is applied, stating that for any integer a, a^5 is congruent to a modulo 5. This implies that the sum of the fifth powers of the integers is congruent to the sum of the integers themselves modulo 5. Since the sum of the integers is zero, it follows that the sum of their fifth powers is also zero modulo 5, confirming divisibility by 5. The key takeaway is that the condition of the integers summing to zero is crucial for the proof.
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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.
 
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Monoxdifly said:
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.
 
castor28 said:
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?
 
Monoxdifly said:
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 $5$.
 
I think I kinda get a grasp here. Thanks for your help.
 
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