If we allow the sum to be a different power ... The sum of the first N cubes is the square of a triangle. 1^3 + 2^3 + 3^3 + 4^3 = 10^2 Sometimes a triangle is also a square: T(8) = 36 = 6^2. The square of a square is a 4th power: 1^3 + 2^3 + ... + 8^3 = 1296 = 36^2 = 6^4 The next triangle-square is 1225. 1^3 + 2^3 + ... + 49^3 = 1225^2 = 35^4 There's a hidden fact in the Ramanujan-Hardy 1729 story, but it's never mentioned. 1729 is a (base 2) pseudoprime: 2^1728 (mod 1729) = 1. Here's how I flunked my Ramanujan test. My University of Arizona employee number was 41616. I didn't look at it very often-- mainly when I used my ID card to check out a library book. Years later, I noticed the number was special. In fact, very special. Puzzle: Why is it special? Puzzle2: Why is it very special? Rich ------- Quoting James Buddenhagen <jbuddenh@gmail.com>:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun