Well, there is 1^1 + 2^1 = 3^1, which I realize does not have k > 2. Closely related to your question, so perhaps of interest even though it is not a match is: http://oeis.org/A030052/a030052.txt Jim On Fri, May 8, 2020 at 2:18 PM Dan Asimov <dasimov@earthlink.net> wrote:
Famously,
1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2
is the only case where the sum of an initial sequence of squares equals another square (ignoring 1^2 = 1^2).
Are there any non-trivial examples of this for powers higher than 2 ?
(For non-initial cases, 3^3 + 4^3 + 5^3 = 6^3 is the first example, and this website <https://www.mathpages.com/home/kmath147.htm> gives many other examples for cubes.)
But I'm interested in when 1^k + 2^k + ... + n^k is an exact kth power for k > 2 and of course n > 1.
—Dan
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