5
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$$ 1^5+2^5+3^5+\dots+14^5+15^5 \pmod{13} $$

I found this question on my old textbook, it seems very trivial but my answer and the answer of book are different. My solution is that:

$$ [1^5+2^5+3^5+4^5+5^5+6^5+(-6)^5+(-5)^5+(-4)^5+(-3)^5+(-2)^5+(-1)^5+0^5+1^5+2^5] \pmod{13} $$

So, it gives us $33\pmod{13}=7$, but the answer is $8$. Where am I missing?

Moreover, if you know any trick or shortcut for these types of problem, can you share your knowledge?

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7
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    $\begingroup$ Wolfram says $7$ too. Your book may be wrong. $\endgroup$
    – Aphelli
    Commented Sep 1, 2020 at 8:18
  • $\begingroup$ I say $7$ too, or $-6$. $\endgroup$
    – Bernard
    Commented Sep 1, 2020 at 8:41
  • $\begingroup$ The correct answer is $7$ $\endgroup$
    – Peter
    Commented Sep 1, 2020 at 8:52
  • $\begingroup$ For a shortcut, use the formula for the sum of the first $n$ fifth powers. But your approach is already short anyway (and elegant!) $\endgroup$
    – Peter
    Commented Sep 1, 2020 at 8:57
  • $\begingroup$ If such tricks do not work, best is to use Faulhabers formula's. $\endgroup$
    – Peter
    Commented Sep 1, 2020 at 9:01

1 Answer 1

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3
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Since $5$ is prime to $\phi(13)=12$, each residue $\bmod 13$ is the fifth power of one residue and then

$1^5+2^5+...13^5\equiv1+2+...+13\equiv 78\equiv0\bmod 13$.

So the given sum reduces to $14^5+15^5\equiv1^5+2^5\equiv33\equiv7$.

And $7$ is correct.

Either the book is wrong or, as sometimes happens, you mistakenly read an answer to an adjacent problem. That, of course, cannot be resolved here.

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10
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    $\begingroup$ Why bother calculating the sum of the first thirteen positive integers when we know $\mathbf Z/13\mathbf Z$ is a group, so the sum is obvious? $\endgroup$
    – Bernard
    Commented Sep 1, 2020 at 10:12
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    $\begingroup$ How do you know that getting the residues to the fifth power is a permutation? $\endgroup$
    – Everstudent
    Commented Sep 1, 2020 at 10:18
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    $\begingroup$ The sum of all elements of a group depends on whether the order of the group is even or odd. $\endgroup$
    – badjohn
    Commented Sep 1, 2020 at 10:19
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    $\begingroup$ Well going Z13 is a group of order 13, so each element has order 13. So how do we know that $a^5 \neq b^5$ $\endgroup$
    – Everstudent
    Commented Sep 1, 2020 at 10:22
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    $\begingroup$ @OscarLanzi: I forgot to add it's a group of odd order. $\endgroup$
    – Bernard
    Commented Sep 1, 2020 at 10:30

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