Splitting the number line so the sum of one side equals the sum of the other.

Factual Questions
1

There are a few instances of this, starting with
1 + 2 = 3

The next occurrence is
Sum (1 to 14) = Sum (15 to 20) = 105

the remaining ones I have found are:
Sum (1 to 84) = Sum (85 to 119) = 28,561

Sum (1 to 492) = Sum (493 to 696) = 121,278

Sum (1 to 2,870) = Sum (2,871 to 4,059) = 4,119,885

Sum (1 to 16,730) = Sum (16,731 to 23,660) = 139,954,815

Sum (1 to 97,512) = Sum (97,513 to 137,903) = 4,754,343,828

Sum (1 to 568,344) = Sum (568,345 to 803,760) = 161,507,735,340

Sum (1 to 3,312,554) = Sum (3,312,555 to 4,684,659) = 5,486,508,657,735

Sum (1 to 19,306,982) = Sum (19,306,983 to 27,304,196) = 186,379,786,627,653

Sum (1 to 112,529,340) = Sum (112,529,341 to 159,140,519) = 6,331,426,236,682,470

One mathematical pattern is that the ‘split’ occurs at 1/sqrt(2)
(e.g. 159,140,519 * 1/sqrt(2) = 112,529,340.15)

To predict the next occurrence, multiply the final number of the previous number line by 5.828…

Well, I know this must have been done by someone before, but I was just curious if anyone knows who it was (Descartes, D’Allembert, Isaac Newton, an SDMB member?)
Does anyone know any history of this?

2

1 + 2 + … + m = m(m + 1)/2, and (m + 1) + (m + 2) + … n = (n - m)(m + n + 1)/2, so all you need to do is set those two expressions equal and solve for n in terms of m. I get n = (\sqrt{8m^2 + 8m + 1} - 1)/2, so for any value of m where that’s an integer, you have a solution.

3

I’ll use the same notation as ultrafilter, but head in a different direction with the algebra:
1 + … + m = (m+1) + … + n
m(m+1) / 2 = n(n+1)/2 - m(m+1)/2
m(m+1) = n(n+1)/2
8m[sup]2[/sup]+8m = 4n[sup]2[/sup]+4n
2(2m+1)[sup]2[/sup] - 2 = (2n+1)[sup]2[/sup]-1
2(2m+1)[sup]2[/sup] - (2n+1)[sup]2[/sup] = 1
and so
2r[sup]2[/sup] - s[sup]2[/sup] = 1
where I have defined r=2m+1 and s=2n+1. This is an instance of one of the most important two-variable quadratic Diophantine equations, Pell’s equation. (The Wikipedia article calls this version a negative Pell equation.)

As the Wikipedia article mentions, solutions to Pell’s equation can be found by continued fractions (in this case, using the continued-fraction convergents to sqrt(2)) or by working in the ring Z[sqrt(2)]. In this case, in particular, you can show that if (r,s) is a solution then so is (3r+2s,4r+3s). Starting with the trivial solution (r,s)=(1,1) you can generate the sequence of solutions
(5,7), (29,41), (169,239), (985,1393), …
and then by taking m=(r-1)/2 and n=(s-1)/2 you get the solutions you want:
(2,3), (14,20), (84,119), (492,696), …

You can also use the expressions for r and s in terms of m and n to find a recurrence relation directly for the pairs (m,n).

4

(You don’t need to start with 1. For example 12+13+14+15+16+17+18 = 19+20+21+22+23)

This pattern is discussed at the Integer Sequences website. It mentions little or no pre-21st century history.

5.828…= 3 + √8

5

Thanks ultrafilter, **Omphaloskeptic **and septimus for those detailed answers.
It’s a little disappointing that this problem doesn’t even have an interesting anecdote such as the Monkey and the Coconuts Puzzle.
Thanks again.