Given the Diophantine equation,


$$x\_1^k+x\_2^k+x\_3^k = y\_1^k+y\_2^k+y\_3^k\tag1$$


there is the rather curious observation that the **smallest** positive solutions for $k=5$ or $6$ is *multi-grade*.


$$24^k+28^k+67^k=3^k+54^k+62^k,\quad k = 1,5$$


$$15^k + 10^k + 23^k = 3^k + 19^k + 22^k,\quad k = 2,6$$


Duncan Moore has exhaustively searched $(1)$ for all positive and **primitive** solutions below a bound $Z$. [Table 1](http://homepage.ntlworld.com/duncan-moore/taxicab/3_3_5.txt) is for $k=5$, while [Table 2](http://homepage.ntlworld.com/duncan-moore/taxicab/3_3_6.txt) is for $k=6$. We summarize the data below.


**I. Table 1:**


$$\begin{array}{|c|c|c||}
\text{# of solns}&\color{blue}{A:=\text{(% of}\; k = 1,5)}&\text{diff}\\
2^0\cdot168&63.7\text{%}& \\
2^1\cdot168&65.8\text{%}&+2.7\\
2^2\cdot168&65.6\text{%}&-0.3 \\
2^3\cdot168&63.6\text{%}&-2.0\\
2^4\cdot168&61.0\text{%}&-2.6\\
2^5\cdot168&59.1\text{%}&-1.9\\
\end{array}$$


*Note:* To address one comment below, $A$ is the percentage of solns given in the first column that is valid for both $k=1,5$. For example, out of the first $2^5\cdot168 = 5376$ solns, then $59.1\text{%}$ are for $k=1,5$.


Each row doubles the $\text{#}$. Since Moore's database has $5393$ solns, and $5393/2^5\approx168.53$, then I used that as the base value. 


**II. Table 2:**


$$\begin{array}{|c|c|c|}
\text{# of solns}&\color{blue}{B:=\text{(% of}\; k = 2,6)}&\text{diff}\\
50&80\text{%}& \\
100&85\text{%}&+5.0\\
200&89\text{%}&+4.0\\
400&91.7\text{%}&+2.7\\
\end{array}$$


*Note:* Thus, out of the first $400$ solns, then a whopping $91.7\text{%}$ of them are actually multi-grade for $k=2,6$. (I'm not sure if excluding non-primitive solutions below the bound $Z$ is relevant. Program-wise, it seems easier to just include them.)


**Questions:**


1. Why is the percentage of $A$ *decreasing*, while that of $B$ is *apparently increasing*? Or will $B$ eventually have a negative diff like $A$? (The data is too small to be conclusive.)
2. If both are decreasing, will $A,B \to 0$? Or will it taper off to some constant?


**P.S.** [This answer](https://mathoverflow.net/a/150430/12905) to a related post might be informative. Incidentally, the smallest solutions to,


$$x\_1^k+x\_2^k+x\_3^k+x\_4^k = y\_1^k+y\_2^k+y\_3^k+x\_4^k\tag2$$


are **also** multigrades as $k=1,5$, and $k=2,6$, though there are no exhaustive tables for these.



Density of multi-grade solutions to $x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k$ for $k = 5$ or $6$?
This is more of a long comment than an answer.


This problem is very difficult and it is not clear to me whether anyone will be able to prove the statments you are looking for. The usual way of studying such equations is to use the circle method, however you have too few variables to get the circle method to work.


It seems quite likely however that this phenomenon is compatible with [Manin's conjecture](http://en.wikipedia.org/wiki/Manin_conjecture). Namely, let 
$$X\_k : \quad x^k\_1+x^k\_2+x^k\_3=y^k\_1+y^k\_2+y^k\_3$$ 
denote the corresponding hypersurface in $\mathbb{P}^5$. You are essentially interested in the rational points of bounded height on $X\_k$ (okay, you also have a positivity condition which corresponds to some condition at the real place, and throwing this away should not change things too much, provided that you also throw away the other trivial solutions with the $x\_i$ or $y\_i$ negative which arise).
In particular how the rational points of bounded height on $X\_5$ (resp. $X\_6$) compares with those on the subvariety $X\_5 \cap X\_1$ (resp. $X\_6 \cap X\_2$). 


An important principle in Manin's conjecture is that there may be "accumulating subvarieties", which may have more (or a comparable number of) solutions than the total space, and one needs to take these into account separately. You are asking whether $X\_5 \cap X\_1$ (resp. $X\_6 \cap X\_2$) is an accumulating subvariety of $X\_5$ (resp. $X\_6$).


The variety $X\_5$ is Fano, and Manin's conjecture predicts that once one removes all accumulating subvarieties, then the number solutions of height less than $Z$ is asymptotic to some constant times $Z$. My guess here is that the accumulating subvarieties exactly correspond to the trivial solutions, hence the subvariety $X\_5 \cap X\_1$ should not be accumulating (i.e. $A \to 0$)


The variety $X\_6$ is Calabi-Yau, so things are much more complicated here. There is a version of Manin's conjecture in this setting, but it is a bit tricky to state. For example it is quite posible that the accummulating subvarieties could be Zariski dense. I don't think your numerical evidence is conclusive to be able to deduce anything here, but it seems to hint that $X\_6 \cap X\_2$ could be accumulating.


The equations which you are interested in are very similar to those which arise in [Vinogradov's mean value theorem](http://en.wikipedia.org/wiki/Vinogradov%27s_mean-value_theorem). One approach would be to try to use recent developments here due to Trevor Wooley to tackle the problem, but I did not investigate this.


