# Splitting the number line so the sum of one side equals the sum of the other. **URL:** \ **Category:** Factual Questions\ **Created:** [April 27, 2013, 12:35am UTC](https://boards.straightdope.com/t/splitting-the-number-line-so-the-sum-of-one-side-equals-the-sum-of-the-other/656809 "2013-04-27T00:35:28Z")\ **Posts on this page:** 5\ **Page:** 1
**Author:** ![wolf\_meister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/wolf_meister/32/15202_2.png) [@wolf\_meister](https://boards.straightdope.com/u/wolf_meister)\ **Post date:** [April 27, 2013, 12:35am UTC](https://boards.straightdope.com/t/splitting-the-number-line-so-the-sum-of-one-side-equals-the-sum-of-the-other/656809/1 "2013-04-27T00:35:28Z")
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? ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [April 27, 2013, 12:44am UTC](https://boards.straightdope.com/t/splitting-the-number-line-so-the-sum-of-one-side-equals-the-sum-of-the-other/656809/2 "2013-04-27T00:44:20Z")
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. ---
**Author:** ![Omphaloskeptic](https://avatars.discourse-cdn.com/v4/letter/o/bcef8e/32.png) [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)\ **Post date:** [April 27, 2013, 3:33am UTC](https://boards.straightdope.com/t/splitting-the-number-line-so-the-sum-of-one-side-equals-the-sum-of-the-other/656809/3 "2013-04-27T03:33:29Z")
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](http://en.wikipedia.org/wiki/Pell%27s_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). ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [April 27, 2013, 4:06am UTC](https://boards.straightdope.com/t/splitting-the-number-line-so-the-sum-of-one-side-equals-the-sum-of-the-other/656809/4 "2013-04-27T04:06:07Z")
(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.](http://oeis.org/A053141) It mentions little or no pre-21st century history. 5.828…= 3 + √8 ---
**Author:** ![wolf\_meister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/wolf_meister/32/15202_2.png) [@wolf\_meister](https://boards.straightdope.com/u/wolf_meister)\ **Post date:** [April 27, 2013, 6:59am UTC](https://boards.straightdope.com/t/splitting-the-number-line-so-the-sum-of-one-side-equals-the-sum-of-the-other/656809/5 "2013-04-27T06:59:01Z")
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.