Is there a better way to finish this number theory/algebra problem other than trial and error?

Question

Problem : In a street, all the houses are numbered continuously from $1$ to $1000$. Alice lives at number $6$. The sum of the numbers from $1$ to $5$ is equal to that of the numbers from $7$ to $8$, i.e. $15$. Then, Bob lives at number $35$. The sum of the numbers from $1$ to $34$ is equal to that of the numbers from $36$ to $49$, i.e. $595$. Finally, Cedric lives at number $N$, different from Alice’s and Bob’s numbers. The sum of the numbers from $1$ to $(N - 1)$ is equal to that of the numbers from $(N + 1)$ to a certain number strictly greater than $(N + 1)$. What is the value of $N$?

I got to the equation $2N^2=K^2+K$ where K is the number of the higher number house and the official solution consists in listing all the perfect squares from $1$ to $1500$ and their doubles in order to find all possible $N$ and $K$ that satisfy the equation. I'm guessing there's a more efficient way to get to the answer, but I couldn't find it...

Answer 1

Yes, there is a way to solve it using a Pell equation.

We want to find $N$ and $K$ such that $$2N^2 = K^2 + K.$$ Multiply by $4$ on both sides and add one to get $$8N^2 + 1 = 4K^2 + 4K + 1.$$ You can rearrange this as $$(2K + 1)^2 - 2(2N)^2 = 1.$$ Compare this to the Pell equation $$x^2 - 2y^2 = 1.$$ It is well known that the pair $(x, y)$ is a solution to this equation if and only if $$x + y\sqrt{2} = (3 + 2\sqrt{2})^n$$ for some integer $n$. Alice's and Bob's houses correspond to the cases $n = 2$ and $n = 3$. (Check this on your own.) Cedric's house corresponds to $n = 4$. Writing it out, $$(2K + 1) + (2N)\sqrt{2} = (3 + 2\sqrt{2})^4 = 577 + 408\sqrt{2},$$ so that $\boxed{(N, K) = (204, 284)}$.

Answer 2

Write it as $8N^2+1=4K^2+4K+1=(2K+1)^2=2(2N)^2+1$ and you have the standard Pell equation. Alice represents the solution $2K+1=17, 2N=12.$ Bob represents the solution $2K+1=99, 2N=70$. The next solution is $2K+1=577, 2N=408$ which can be found by the standard recursion $(2K+1)'=3(2K+1)+4(2N), (2N)'=2(2K+1)+3N$, so Cedric's house number is $204$ and the higher number is $288$.