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

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

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?

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

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.

• 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? Commented Sep 1, 2020 at 10:12
• How do you know that getting the residues to the fifth power is a permutation? Commented Sep 1, 2020 at 10:18
• The sum of all elements of a group depends on whether the order of the group is even or odd. Commented Sep 1, 2020 at 10:19
• Well going Z13 is a group of order 13, so each element has order 13. So how do we know that $a^5 \neq b^5$ Commented Sep 1, 2020 at 10:22
• @OscarLanzi: I forgot to add it's a group of odd order. Commented Sep 1, 2020 at 10:30