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I´m asked to find the remainder of dividing $\sum_{i=0}^{1080}i^5$ by $14$. How can I do this using only basic results from modular arithmetic? Only one thing comes to my mind, here's my idea: we know that each number has a representative $r$ in the class of $\pmod{14}$, that satisfies $0\leq r<14$, so in order to simplify things, it is only needed to find this representative for $n^5$ for the integers $n$ between $0$ and $13$, because, for example, if I wanted to find the residue of $\sum_{i=0}^{27}i^5$ divided by $14$, then assuming I know that $k_n$ is the representative of $n$ for each $n\in\{ 0,\dots,13 \}$, then $14\equiv0\pmod{14}$ implies $14^5\equiv0^5\equiv k_0\pmod{14}$, $15\equiv1\pmod{14}$ implies $15^5\equiv1^5\equiv k_1\pmod{14}$, and so on until $27^5\equiv13^5\equiv k_{13}\pmod{14}$. In this way, it is now possible to know that, given that \begin{equation*}\sum_{i=0}^{27}i^5=\sum_{i=0}^{13}i^5+(i+14)^5 \end{equation*}, then for $i\in\{ 0,\dots,13 \}$, $i^5\equiv (i+14)^5\pmod{14}\Rightarrow i^5+(i+14)^5\equiv2i^5\equiv2k_i\pmod{14}$, which implies that: \begin{equation*}\sum_{i=0}^{13}i^5+(i+14)^5\equiv\sum_{i=0}^{13}2k_i\equiv R_k\pmod{14} \end{equation*} where $R_k$ is the representative of $\sum_{i=0}^{13}2k_i$ in $\mathbb{Z}_{14}$. That's the residue we're looking for.

Following the pattern, I'd need to find out how many times a number which is congruent to a number between $0$ and $13$ in $\mathbb{Z}_{14}$ appears between $0$ and $1080$, and then rewrite the original sum in terms of how many times the "repeated" (by this I mean in the sense that they are equivalent in $\mathbb{Z}_{14}$) numbers appear.

(I already did this, but as I said, it's a very long and tedious process).

Another idea is to use the formula for the sum of the first $n$ fifth powers.

Any other idea of an easier process, or a check to mine would be really appreciated. Thanks in advance.

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    $\begingroup$ By the Chinese Remainder Theorem you only need to consider the congruence mod $7$ and mod $2$. $\endgroup$ – Lukas Kofler Apr 18 '20 at 23:54
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The mapping $x\mapsto x^5$ is a bijection on integers modulo $14$ (its inverse is itself),

and $1080=1078+2=77\times14+2$.

Therefore, $\sum\limits_{i=0}^{1080}i^5\equiv\sum\limits_{i=0}^{1077}i+1078^5+1079^5+1080^5$

$\equiv77\sum\limits_{i=0}^{13}i+0^5+1^5+2^5\equiv77\times\dfrac{13\times14}2+1+32$

$\equiv7\times odd+1+32\equiv7+33=40\equiv12\bmod14.$

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  • $\begingroup$ How can you prove it's bijective? I've been trying and I just can't seem to get anywhere. $\endgroup$ – Bryan Castro Apr 19 '20 at 17:54
  • $\begingroup$ One way would be to compute $0^5, (\pm1)^5, (\pm2)^5, (\pm3)^5, (\pm4)^5, (\pm5)^5, (\pm6)^5, $ and $7^5$ mod $14$; another way, which I alluded to, is to show that $n^7\equiv n\bmod 2$ and $7$ so $\bmod 14$, and therefore $(n^5)^5=n^{25}=(n^7)^3n^4\equiv n^3n^4\equiv n^7\equiv n\bmod14$, so $n\mapsto n^4$ is bijective because it has an inverse $\endgroup$ – J. W. Tanner Apr 19 '20 at 19:00
  • $\begingroup$ I made a typo in my comment above; I meant $n\mapsto n^5$ where I typed $n\mapsto n^4$ $\endgroup$ – J. W. Tanner Apr 19 '20 at 20:21
  • $\begingroup$ Hmm, that seems very good. Thank you very much! I would upvote your answer but I'm not able until I get 15 reputation. Greetings. $\endgroup$ – Bryan Castro Apr 20 '20 at 1:17
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We have

$\tag 1 \displaystyle \sum_{i=0}^{13}i^5 \equiv 0 + + 7^5 +\sum_{i=1}^{6}\bigr(i^5+(-i)^5\bigr) \equiv 7^5 \equiv 7 \pmod{14} $

Also, since

$\quad 1081=77\times14+3$ we can write

$ \tag 2 \displaystyle \sum_{i=0}^{1080}i^5 \equiv 77\cdot7 + 0^5 + 1^5 + 2^5 \equiv 7 + 0 + 1 + 4 \equiv 12\pmod{14} $

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