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

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Discussion Overview

The discussion revolves around demonstrating that the sum of the fifth powers of ten integers, which sum to zero, is divisible by 5. The scope includes mathematical reasoning and the application of Fermat's Little Theorem.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about how to approach the problem, suggesting brute-force methods may be impossible.
  • Another participant introduces Fermat's Little Theorem, stating that for any integer \(a\), \(a^5 \equiv a \pmod{5}\), and suggests this theorem is applicable regardless of the number of integers.
  • A participant questions whether the condition of the sum being zero is merely a distraction, seeking clarification on the relevance of the number of integers.
  • It is clarified that while the number of integers is not crucial, the condition that their sum is zero is essential for the proof.
  • The application of Fermat's theorem is elaborated, showing that the sum of the fifth powers of the integers is congruent to the sum of the integers themselves modulo 5, leading to the conclusion that the sum of the fifth powers is divisible by 5.
  • One participant indicates they are beginning to understand the explanation provided.

Areas of Agreement / Disagreement

Participants appear to agree on the application of Fermat's Little Theorem and the importance of the sum being zero, but there is some initial uncertainty regarding the relevance of the number of integers involved.

Contextual Notes

The discussion does not resolve whether the initial participant's approach of brute-forcing is valid or necessary, and the implications of the number of integers remain somewhat ambiguous.

Monoxdifly
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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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