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

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I believe: sum(i=1,n)i^3 = (n*(n+1)/2)^2 Trivially true for n=1. And if true up to n, then: sum(i=1,n+1)i^3 = sum(i=1,n)i^3 + (n+1)^3 = (n*(n+1)/2)^2 + (n+1)^3 = (1/4)*(n*(n+1))^2 + (n+1)^3 = (1/4) * ((n*(n+1))^2 + 4*(n+1)^3) = (1/4) * (n^2*(n+1)^2 + 4*(n+1)*(n+1)^2) = (1/4) * (n^2 + 4*(n+1))*(n+1)^2 = (1/4) * (n+2)^2*(n+1)^2 = ((n+1)*(n+2)/2)^2 So it's true for n+1. Tom christopher landauer writes:

hihi -

i was under the impression that an initial sum of cubes is always a square - isn't there a summation formula to that effect?

more later,

chris

On 2020-05-09 14:09, rcs@xmission.com wrote:

...

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

(But not consecutive) On Sat, May 9, 2020 at 3:35 PM Mike Stay <metaweta@gmail.com> wrote:

There are infinitely many solutions to a^4 + b^4 + c^4 + d^4 = (a + b + c + d)^4 https://en.wikipedia.org/wiki/Jacobi%E2%80%93Madden_equation

On Fri, May 8, 2020 at 1:18 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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-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com