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$$\begin{align} 2^{\color{pink}0}+7^{\color{pink}0}+8^{\color{pink}0}+18^{\color{pink}0}+19^{\color{pink}0}+24^{\color{pink}0}&=3^{\color{pink}0}+4^{\color{pink}0}+12^{\color{pink}0}+14^{\color{pink}0}+22^{\color{pink}0}+23^{\color{pink}0}\\ 2^{\color{red}1}+7^{\color{red}1}+8^{\color{red}1}+18^{\color{red}1}+19^{\color{red}1}+24^{\color{red}1}&=3^{\color{red}1}+4^{\color{red}1}+12^{\color{red}1}+14^{\color{red}1}+22^{\color{red}1}+23^{\color{red}1}\\ 2^{\color{orange}2}+7^{\color{orange}2}+8^{\color{orange}2}+18^{\color{orange}2}+19^{\color{orange}2}+24^{\color{orange}2}&=3^{\color{orange}2}+4^{\color{orange}2}+12^{\color{orange}2}+14^{\color{orange}2}+22^{\color{orange}2}+23^{\color{orange}2}\\ 2^{\color{green}3}+7^{\color{green}3}+8^{\color{green}3}+18^{\color{green}3}+19^{\color{green}3}+24^{\color{green}3}&=3^{\color{green}3}+4^{\color{green}3}+12^{\color{green}3}+14^{\color{green}3}+22^{\color{green}3}+23^{\color{green}3}\\ 2^{\color{blue}4}+7^{\color{blue}4}+8^{\color{blue}4}+18^{\color{blue}4}+19^{\color{blue}4}+24^{\color{blue}4}&=3^{\color{blue}4}+4^{\color{blue}4}+12^{\color{blue}4}+14^{\color{blue}4}+22^{\color{blue}4}+23^{\color{blue}4}\\ 2^{\color{brown}5}+7^{\color{brown}5}+8^{\color{brown}5}+18^{\color{brown}5}+19^{\color{brown}5}+24^{\color{brown}5}&=3^{\color{brown}5}+4^{\color{brown}5}+12^{\color{brown}5}+14^{\color{brown}5}+22^{\color{brown}5}+23^{\color{brown}5}\\ \end{align}$$

Are there similar examples? Any generalization?

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  • 1
    $\begingroup$ Wow, for the first look even this is impossible... $\endgroup$ – Jihad Jan 4 '15 at 14:46
  • 3
    $\begingroup$ You could even add a row for the $0$th powers... $\endgroup$ – user133281 Jan 4 '15 at 14:52
  • $\begingroup$ How did you find this? $\endgroup$ – Alex Silva Jan 4 '15 at 15:02
  • $\begingroup$ @AlexSilva in tumblr $\endgroup$ – user153330 Jan 4 '15 at 15:03
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Finding patterns like this is known as the Prouhet-Tarry-Escott problem, see Wikipedia, also Chen, also Piezas.

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For those who want the quick version, a particular example of Theorem 5 in the link cited by Zander states that if,

$$a^k+b^k+c^k = d^k+e^k+f^k$$

for $k=2,4$, then,

$$\small(x+a)^k+(x+b)^k+(x+c)^k+(x-a)^k+(x-b)^k+(x-c)^k = \\ \small (x+d)^k+(x+e)^k+(x+f)^k+(x-d)^k+(x-e)^k+(x-f)^k$$

for $k=1,2,3,4,5$. The example by the OP used,

$$5^k + 6^k + 11^k = 1^k + 9^k + 10^k$$

and $x=13$. However, to add a nice twist to this post, note the $6-10-8$ identity,

$$64\big(5^6 + 6^6 + 11^6 -(1^6 + 9^6 + 10^6)\big)\big(5^{10} + 6^{10} + 11^{10} -(1^{10} + 9^{10} + 10^{10})\big) =\\ 45\big(5^8 + 6^8 + 11^8 -(1^8 + 9^8 + 10^8)\big)^2$$

To know why, see this MO post.

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Define \begin{align} x\otimes y=x^2+xy+y^2. \end{align} Take integers $t_2>t_3>0, u_2>u_3>0$ with $h_t\ne h_u$ where $h_\alpha=\alpha_2\otimes \alpha_3$. Let $\alpha_1=-\alpha_2-\alpha_3$ and \begin{align} a_1&=t_3u_3-t_1u_1\notag\\ a_2&=t_1u_1-t_2u_2\notag\\ a_3&=t_2u_2-t_3u_3\notag\\ b_1&=t_2u_3-t_1u_1\notag\\ b_2&=t_1u_1-t_3u_2\notag\\ b_3&=t_3u_2-t_2u_3\notag\\ a_{7-i}&=-a_i\tag{asym}\label{asym}\\ b_{7-i}&=-b_i\tag{bsym}\label{bsym} \end{align} Then, for $j=0,\dots,5$ and $n=6$, \begin{align} \sum_{i=1}^n a_i^j=\sum_{i=1}^n b_i^j=\Sigma_j,\tag{te}\label{te} \end{align} say. $\Sigma_0=n=6$, and \ref{asym} and \ref{bsym} ensure $\Sigma_1=\Sigma_3=\Sigma_5=0$. Moreover, \begin{align} \Sigma_2=4h_t h_u; \Sigma_4=4(h_t h_u)^2 \implies\frac{\Sigma_4}{\Sigma_2^2}&=\frac14.\tag{reg}\label{reg} \end{align} \eqref{reg} is not a necessary precondition for \eqref{te}. For example, $$a_{1\dots3}=25, 23, 6; b_{1\dots3}=27, 19, 10$$ satisfy \eqref{te} with $\Sigma_2=1190$ and $\Sigma_4=671762$. However, \eqref{reg} is satisfied for all of the first 18 examples (ordered by increasing $\Sigma_2$) and most of the first few thousand examples.

$$a_2\otimes a_3=b_2\otimes b_3=h_t h_u.$$ Such an equality implies that, for each of the above $\otimes$ formulas in turn, its operands may be used as the values of $t_2, t_3$ with another pair of values for $u_2, u_3$ to produce a collection of up to four sets of six values each, where all those sets of six have the same $\Sigma_2$ and the same $\Sigma_4$, so any two of those sets of six satisfy \eqref{te} as the $a_i$ and $b_i$. For example, $$\begin{array}{rrrrrrrrrrrrr} t_2&t_3&u_2&u_3&h_t&h_u&h_t h_u&a_1&a_2&a_3&b_1&b_2&b_3\\ 2&1&3&1&7&13&91&-11&\color{red}6&\color{red}5&-10&\color{blue}9&\color{blue}1\\ \color{red}6&\color{red}5&3&2&91&19&1729&-45&37&8&-43&40&3\\ \color{blue}9&\color{blue}1&3&2&91&19&1729&-48&23&25&-32&47&-15 \end{array}$$

\eqref{te} may also be stated as \begin{align} \sum_i f(a_i)=\sum_i f(b_i)\tag{poly}\label{poly} \end{align} for $f=x\mapsto x^j$ for $j=0,\dots,k$. This implies that \eqref{poly} is true for $f$ being any polynomial function of degree $\leqslant k$. In particular, we may take $f=x\mapsto (x+d)^j$. Thus if a constant $d$ is added to each of the $a_i$ and $b_i$ in any example, the result also satisfies \eqref{te}. The above table's top row gives $$a_{1\dots 6}=-11, 6, 5, -5, -6, 11; b_{1\dots 6}=-10, 9, 1, -1, -9, 10.$$ Re-ordering and using $d=13$ yields the OP's example.

The problem of finding disjoint sequences $a_1,\dots,a_n$ and $b_1,\dots,b_n$ for $n$ as small as possible, satisfying \eqref{te} for $j=0,\dots,k$, is the Tarry-Escott problem. Prouhet's name is often associated with this problem, but this is wrong; Prouhet's problem is, for $n$ as small as possible, to partition the set $\{0,\dots,n-1\}$ into two or more sets, any two of which satisfy \eqref{te} for $j=0,\dots,k$.

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  • $\begingroup$ thanks fo rtaking interest in this question and thanks for the efforts you put into this answer :))) $\endgroup$ – user153330 Sep 30 '18 at 12:57

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