21
$\begingroup$

As far as I understand, both of the Diophantine equations $$a^5 + b^5 = c^5 + d^5$$ and $$a^6 + b^6 = c^6 + d^6$$ have no known nontrivial solutions, but $$24^5 + 28^5 + 67^5 = 3^5+64^5+62^5$$ and $$3^6+19^6+22^6 = 10^6+15^6+23^6$$ among many other solutions are known, when the number of summands is increased from $2$ to $3$. My information here is at least a decade out of date, and I am wondering if the resolution of Fermat's Last Theorem has clarified this situation, with respect to sums of an equal number of powers...?

$\endgroup$
0

2 Answers 2

21
$\begingroup$

The short answer to your specific question is no, the resolution of FLT via modularity of elliptic curves does not seem to be helpful in dealing with rational points on higher dimensional varieties. The first two equations you list, $a^5+b^5=c^5+d^5$ and $a^6+b^6=c^6+d^6$, are surfaces of general type in $\mathbb{P}^3$, so conjectures of Bombieri, Lang, Vojta, would say that their solutions, and solutions of similar equations in which the terms might have non-unit coefficients, lie on finitely many curves. The second two equations are 4-folds in $\mathbb{P}^5$. For $x^5+y^5+z^5=u^5+v^5+w^5$, the canonical bundle is ample, and I'm going to guess that the 4-fold is rational(?), so there will be lots of solutions. For $x^6+y^6+z^6=u^6+v^6+w^6$, the canonical bundle is trivial, so you've got a Calabi-Yau. At least conjecturally, there should be a number field $K$ such that the $K$-rational points are Zariski dense. Maybe for this 4-fold, one can take $K=\mathbb{Q}$?

$\endgroup$
2
  • 6
    $\begingroup$ Thanks! Yes, there are in infinite number of solutions to the "5.3.3" equation $x^5+y^5+z^5=u^5+v^5+w^5$, going back to Swinnerton-Dyer in 1952. $\endgroup$ Dec 1, 2013 at 0:23
  • 4
    $\begingroup$ Andrew Bremner has written papers on the two 6-term equations. MR0634204 (83g:14017) A geometric approach to equal sums of fifth powers, J. Number Theory 13 (1981), no. 3, 337–354 and MR0635569 (83g:14018) A geometric approach to equal sums of sixth powers, Proc. London Math. Soc. (3) 43 (1981), no. 3, 544–581. The reviews have links to more recent work. $\endgroup$ Dec 1, 2013 at 22:23
10
$\begingroup$

Label the equation,

$$x_1^k+x_2^k+\dots+x_m^k = y_1^k+y_2^k+\dots+y_n^k$$

as a $(k,m,n)$. Let type of primitive solutions be polynomial identity $P(n)$, or elliptic curve $E$. Then results (mostly) for the balanced case $m=n$ are,

I. Table 1 $$\begin{array}{|c|c|c|} (k,m,n)& \text{# of known solutions}& \text{Type}\\ 3,2,2& \infty&P(n)\\ 4,2,2& \infty&P(n),E\\ 5,3,3& \infty&P(n),E\\ 6,3,3& \infty&P(n),E\\ 7,4,4& \text{many} &E\,?\\ 7,4,5& \infty&P(n)+E\\ 8,4,4& 1&-\\ 9,5,5& \text{many}&-\\ 9,6,6& \infty & E\\ 10,5,5& 0&-\\ \end{array}$$

Note: For $(7,4,5)$, see this MSE answer.

II. Table 2. (For multi-grades) $$\begin{array}{|c|c|c|} (k,m,n)& \text{# of known solutions}&\text{Type}\\ 5,4,4& \infty&P(n),E\\ 6,4,4& \infty&P(n),E\\ 7,5,5& \infty&P(n),E\\ 8,5,5& \infty&E\\ 9,6,6& \text{many}&E\,?\\ 10,6,6& \infty&E\\ 11,7,7& 0&-\\ 12,7,7& 0&-\\ \end{array}$$

Note: A multi-grade is simultaneously valid for multiple $k$. For example, the $(9,6,6)$ is for $k=1,3,5,7,9$ while the $(10,6,6)$ is for $k=2,4,6,8,10$.

$\endgroup$
5
  • $\begingroup$ The whole subject of "equal sums of like powers" is much intriquing, especially the involvement of elliptic curves. In lack of a list of references for the results (does such a list exist?) I have added the $(9,6,6)$ solution of Bremner and Delorme, Math. Comp. 79, 2010, for the sake of completeness, of which I understand you are already well aware. $\endgroup$ Jan 11, 2015 at 18:39
  • $\begingroup$ @JesperPetersen: Ah, yes, the $k=1,3,9$ by Bremner and Delorme. The smallest example was found by Lander et al in 1967, but it was only recently that they realized it was a point in an elliptic curve. Which makes me hopeful for those others that have been labeled "many" that eventually an $E$ will be found for them as well. P.S. Randy Ekl made a list in a paper almost two decades ago so it is outdated. $\endgroup$ Jan 11, 2015 at 18:44
  • 2
    $\begingroup$ Just for info, there is a near-balanced case equivalent to a $(7,0,9)$ by Choudhry that is a hybrid: a polynomial $P(n)$ with coefficients $c_i$ determined by an $E$. However, the $c_i$ - drumroll, pls - are as high as $\approx 10^{1185}$. See this MSE answer. $\endgroup$ Jan 11, 2015 at 21:16
  • $\begingroup$ @TitoPiezasIII: $10^{1185}$---That's an impressive number! $\endgroup$ Jan 12, 2015 at 0:22
  • $\begingroup$ @JosephO'Rourke: Compared to $a^2+b^2=c^2$, a 7th deg polynomial identity that uses coefficients in the range $10^{1185}$ does seem a bit excessive. But it is the smallest in the family. :) $\endgroup$ Jan 12, 2015 at 0:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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