What was Ramanujan's solution?

Solution 1:

An expansion of Andre's comment into a detailed exposition can be found at http://www.johnderbyshire.com/Opinions/Diaries/Puzzles/2009-06.html. It's also at http://www.angelfire.com/ak/ashoksandhya/winners2.html#PUZZANS4.

Solution 2:

We will use

Theorem 1: If $r\gt0$ is irrational and $\left|\,\frac pq-r\,\right|\le\frac1{2q^2}$ where $(p,q)=1$, then $\frac pq$ is a continued fraction convergent for $r$.

and

Theorem 2: If $\left\{\frac{p_k}{q_k}\right\}$ is the sequence of continued fraction convergents for an irrational $r\gt0$, then $r$ is between $\frac{p_k}{q_k}$ and $\frac{p_{k+1}}{q_{k+1}}$ and $\frac{p_{k+1}}{q_{k+1}}-\frac{p_k}{q_k}=\frac{(-1)^k}{q_{k+1}q_k}$.

The problem can be written as $$ \overbrace{\ \frac{x^2-x}2\ }^{\substack{\text{sum of $1$}\\\text{to $x-1$}}}=\overbrace{\frac{n^2+n}2-\frac{x^2+x}2}^\text{sum of $x+1$ to $n$}\tag1 $$ which is equivalent to $$ 8x^2=(2n+1)^2-1\tag2 $$ From $(2)$, we get $\frac{2n+1}x=\sqrt{8+\frac1{x^2}}$; thus, we want an over-estimate for $\sqrt8$ that has an odd numerator. Furthermore, $(2)$ implies $$ \begin{align} \frac{2n+1}x-\sqrt8 &=\sqrt{8+\tfrac1{x^2}}-\sqrt8\\[9pt] &=\frac{1/x^2}{\sqrt{8+\frac1{x^2}}+\sqrt8}\\ &\le\frac1{2\sqrt8\,x^2}\tag3 \end{align} $$ Theorem 1 and inequality $(3)$ say that $\frac{2n+1}x$ needs to be a continued fraction approximant for $\sqrt8$ to satisfy $(2)$.

The continued fraction for $\sqrt8$ is $(2;1,4,1,4,1,\dots)$. The convergents are $$ \begin{array}{c|cc} k&0&\color{#C00}{1}&2&\color{#C00}{3}&4&\color{#C00}{5}&\dots\\\hline \frac{p_k}{q_k}&\frac21&\color{#C00}{\frac31}&\frac{14}5&\color{#C00}{\frac{17}6}&\frac{82}{29}&\color{#C00}{\frac{99}{35}}&\dots \end{array}\tag4 $$ The repeating continued fraction gives the following numerator/denominator recurrence $$ \begin{align} a_{2n}&=4a_{2n-1}+a_{2n-2}\tag{5a}\\ a_{2n+1}&=a_{2n}+a_{2n-1}\tag{5b} \end{align} $$ Theorem 2 says that the red terms, the ones with odd indices, are over-estimates. Furthermore, the recurrences $\text{(5a)}$ and $\text{(5b)}$ guarantee that the red terms have odd numerators.

Combining $\text{(5a)}$ and $\text{(5b)}$, we get that the recurrence for the odd numbered numerators and denominators is $$ a_{2n+1}=6a_{2n-1}-a_{2n-3}\tag6 $$ Thus, we can get the solutions from the odd-indexed terms of $(4)$: $$ \begin{array}{c|cc} k&1&3&5&7&9&\dots\\\hline \frac{p_k}{q_k}&\frac31&\frac{17}6&\frac{99}{35}&\frac{577}{204}&\frac{3363}{1189}&\dots\\ \frac{n_k}{x_k}&\frac11&\frac86&\frac{49}{35}&\frac{288}{204}&\frac{1681}{1189}&\dots \end{array}\tag7 $$ The pair that has $50\le n\le500$, is $(x,n)=(204,288)$.

Checking: $$ \sum_{k=1}^{203}k=20706=\sum_{k=205}^{288}k\tag8 $$

Solution 3:

See Pi Mu Epsilon Journal Vol. 14 # 6 pp 389-97, 2017.
"House Number Puzzle: Solution Through Bifurcation"

A summary from the above Pi Mu Epsilon publication is given below:

Continued Fraction of √2 is: 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169 ... The above (x/y) also form solutions to the Pell's equation: x ² −2 y ² = ±1. The house numbers are products of x and y (the solutions to the above Pell's equation): 1*1; 3*2; 7*5; 17*12; 41*29 ... and the total number of houses are provided by x*x and 2*y*y alternatively. Total number of houses: 1*1; 2*2*2; 7*7; 2*12*12; 41*41 ...

For details see the actual publication.