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 is for $k=5$, while Table 2 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.)


  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 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.

  • $\begingroup$ Can you please clarify the definitions of $A$ and $B$? It is not clear to me at the moment how these are defined. $\endgroup$ – Daniel Loughran Jan 3 '15 at 6:35
  • $\begingroup$ @DanielLoughran: Ok. For $B$, as an example, inspecting the first $50$ solutions of Moore's database for $k=6$, then $80\text{%}$ of them are actually valid for both $k=2,6$. If we inspect the first $100$ solns, then $85\text{%}$ of them are so valid. And so on. $\endgroup$ – Tito Piezas III Jan 3 '15 at 6:47
  • $\begingroup$ You have not said so, however it appears that you are not counting "trivial solutions", e.g. when $x_i = y_i$ for all $i$. Could you please clarify which solutions you are counting? Are there are other solutions which you not count? $\endgroup$ – Daniel Loughran Jan 3 '15 at 10:36
  • $\begingroup$ Have you checked whether some of the non trivial solutions are also valid for other $k$'s between 1 and 6 (or maybe even a bit bigger)? It would be even more puzzling if there is a special 'link' between 1,5 and between 2,6. $\endgroup$ – Wolfgang Jan 3 '15 at 11:10
  • $\begingroup$ @DanielLoughran: I'm basing the counts purely on Moore's database so you are correct, I'm not counting a) "trivial solutions" ($x_i=y_i$); and also not counting b) "non-primitives" (those with all terms having a common factor). $\endgroup$ – Tito Piezas III Jan 3 '15 at 17:39

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. 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. 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.

| cite | improve this answer | |
  • $\begingroup$ 1. Thanks for translating. I've answered your comment above (re the counts). Ah, so there is a big difference between $X_5$ and $X_6$! The first is Fano and the other is Calabi-Yau. That is one of the things I'm looking for: additional info to better understand those equations. By the way, I just checked the database: the set $X_6 \cap X_1$ is empty. You'd think that by coincidence there should be at least a few, but it's zero.$\\$ 2. If we go higher, $$Y_k : x_1^k+x_2^k+x_3^k+x_4^k = y_1^k+y_2^k+y_3^k+y_4^k$$ are there terms to distinguish the varieties $Y_7$ and $Y_8$? $\endgroup$ – Tito Piezas III Jan 3 '15 at 18:09
  • $\begingroup$ Yes something similar happens for the $Y_i$. This is a general geometric fact which follows from the adjunction formula. Namely, let $X$ be a smooth hypersurface of degree $d$ in $\mathbb{P}^n$. If $d < n + 1$, then $X$ is Fano. If $d = n + 1$ then $X$ is Calabi-Yau. If $d > n + 1$ then $X$ is of general type. $\endgroup$ – Daniel Loughran Jan 3 '15 at 18:27
  • $\begingroup$ I've already checked $Y_7$ before. Still with positive terms, a significant portion ($>25\text{%}$) also have $Y_7 \cap Y_1$, and some even have $Y_7 \cap Y_3$ as well. For $Y_8$, only one primitive is known, so nothing much can be said. P.S. Since $X_6 \cap X_1$ is zero in the database, is there an obstruction not present in $X_5 \cap X_1$? $\endgroup$ – Tito Piezas III Jan 3 '15 at 18:38
  • $\begingroup$ I don't know why you can't find solutions for $X_6 \cap X_1$. There could be local obstructions coming from $p$-adic conditions, or maybe the local conditions are satisfied but the Hasse principle fails. Alternatively there could indeed be a solution, but it has extremely large height so you won't ever find it (this is not uncommon for Calabi-Yau varieties and varieties of general type - e.g. see Elkies work on the Euler quartic). $\endgroup$ – Daniel Loughran Jan 3 '15 at 19:06
  • $\begingroup$ I've studied $k=1,2,6$ before, but not with the positivity requirement. The complete solution can be given as, $$(a+b)^k+(c-d)^k+(e+f)^k = (-a+b)^k+(c+d)^k+ (-e+f)^k\tag1$$ where $$d,\;e,\;f = -an,\; -a(n+1),\; (b+nc)/(n+1)$$ This satisfies $k=1,2$ and expanding this to $k=6$ results in only a quadratic in $a$. Rational solutions can be found if its discriminant (only a quartic in $b$) is made a square. One can then transform it to an elliptic curve. I re-visited my old solutions and unfortunately, I can't get the addends $x_i$ of $(1)$ to be all positive. $\endgroup$ – Tito Piezas III Jan 3 '15 at 20:39

Not an answer, just a graphic illustrating the percentages $A$ and $B$ of multi-grade solutions vs. all solutions, sorted according to the biggest contained term. It is hard to believe they should drop to $0$ as $n$ grows, especially $B$, but sure enough, the available data after all only cover a tiny part of infinity... enter image description here

| cite | improve this answer | |
  • $\begingroup$ Thanks for the graphs. But one can always use Skewes' number as a cautionary tale. The smallest known crossing is $\approx e^{728}$. On the other hand, the Birch and Swinnerton-Dyer conjecture in the 1960s initially used rather tenuous trends in graphical plots. Supposedly this made Birch's Ph.D. advisor J. Cassels skeptical, but over time the numerical evidence stacked up. So in the absence of anything definite, it shouldn't hurt to look at more data, especially for $k=6$. $\endgroup$ – Tito Piezas III Jan 4 '15 at 16:38
  • $\begingroup$ Yes, I fully agree :( Thank you that the caution didn't prevent you from posting your question! $\endgroup$ – Wolfgang Jan 4 '15 at 16:59

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.