TITLE: State of knowledge of $a^n+b^n=c^n+d^n$ vs. $a^n+b^n+c^n=d^n+e^n+f^n$
QUESTION [21 upvotes]: 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My information here is at least a decade out of date,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...?

REPLY [10 votes]: 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 $k=2,4,6,8,10$.
{"set_name": "stack_exchange", "score": 21, "question_id": 150428}