Six of a kind .

Question

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

Are there similar examples? Any generalization?

Answer 1

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

Answer 2

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.