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

Are there similar examples? Any generalization?

  • 1
    $\begingroup$ Wow, for the first look even this is impossible... $\endgroup$
    – Jihad
    Jan 4, 2015 at 14:46
  • 3
    $\begingroup$ You could even add a row for the $0$th powers... $\endgroup$
    – user133281
    Jan 4, 2015 at 14:52
  • $\begingroup$ How did you find this? $\endgroup$
    – Alex Silva
    Jan 4, 2015 at 15:02
  • $\begingroup$ @AlexSilva in tumblr $\endgroup$
    – user153330
    Jan 4, 2015 at 15:03

3 Answers 3


Finding patterns like this is known as the Prouhet-Tarry-Escott problem, see Wikipedia, also Chen, also Piezas.

  • $\begingroup$ The above link for Chen is not updated. Here is my new website: eslpower.org $\endgroup$ Aug 25, 2021 at 13:57

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.


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 $j=0,\dots,k$.

  • $\begingroup$ thanks fo rtaking interest in this question and thanks for the efforts you put into this answer :))) $\endgroup$
    – user153330
    Sep 30, 2018 at 12:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.