Infinitely pairs of positive integers such that $1+2+\cdots+k = (k+1)+(k+2)+\cdots+N$

Question

Prove that there are infinitely many pairs $(k,N)$ of positive integers such that $1+2+\cdots+k = (k+1)+(k+2)+\cdots+N$.

I thought of transforming this into a Pell equation, but I didn't see a way of doing that. Should we show that if the equation is true for some $(k,N)$ it is also true for another?

@lulu In that question I asked about a solution. In this one I ask for a solution. — user19405892, Jul 19 '16 at 16:46
@AndréNicolas Suppose that $(k,N)$ is a solution. Then we have $$1+2+\cdots+k = \dfrac{k(k+1)}{2}= (k+1)+(k+2)+\cdots+N = \dfrac{(N-k)(N+k+1)}{2},$$ which means $2k(k+1) = N(N+1)$. This can be transformed into the negative Pell equation $(2N+1)^{2}-2(2k+1)^{2}=-1$. This was the answer given in the other linked question, but I don't know where to go from here. — user19405892, Jul 19 '16 at 16:56
From a solution you get another by "multiplying" by a fundamental solution of the Pellian with right-hand side $1$. — André Nicolas, Jul 19 '16 at 16:59

Riccardo.Alestra · Answer 1 · 2016-07-19T17:12:29.623

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Obviously, because the solution respect to, for example to $k$ is: $$k=N-\dfrac{1}{2}+\dfrac{1}{2}\sqrt{(8N^2+1)},$$ excluding the negative ones, you can get infinite solutions. Given $N$ you can get $k$

edited Jul 19 '16 at 17:12

answered Jul 19 '16 at 16:58

Riccardo.Alestra

10,546

Which equation did you solve to get that? – user19405892 Jul 19 '16 at 17:02
@user19405892: the equation yoy have written in the question. – Riccardo.Alestra Jul 19 '16 at 17:05
I get $k = -\dfrac{1}{2}+\dfrac{1}{2}\sqrt{2 N^2+2 N+1}$. – user19405892 Jul 19 '16 at 17:06
@user19405892: the first sum is $\dfrac{1}{2}(k^2+k)$. The second one is: $kN+\dfrac{1}{2}(N^2+N)$ – Riccardo.Alestra Jul 19 '16 at 17:14
We have $\dfrac{k(k+1)}{2} = \dfrac{1}{2}(k^2+k)$ and $\dfrac{(N-k)(N+k+1)}{2} = \dfrac{N^2+N-k-k^2}{2}$. – user19405892 Jul 19 '16 at 17:16
@user19405892: you have to calculate $\sum_{n=1}^N(k+n)$ – Riccardo.Alestra Jul 19 '16 at 17:21
That's not correct. It goes up to $N$, not $k+N$. – user19405892 Jul 19 '16 at 17:26

score 0 · Answer 2 · edited Jun 12 '20 at 10:38

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$$1+2+\cdots+k = (k+1)+(k+2)+\cdots+N$$ you can find the sums of two sides and you will get a second degree equation solve it and you get: $$k=\frac{-1}{2}+\frac{1}{2}\sqrt{2N^2+2N+1}$$

use this "Maple" instructions you get the solution:

$f:=k->\sum_{i=0}^{k}i-\sum_{j=k+1}^{N}j$:

$solve\left(f(k)=0,k\right)$;

edited Jun 12 '20 at 10:38

Community

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answered Jul 19 '16 at 17:30

TH3

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Infinitely pairs of positive integers such that $1+2+\cdots+k = (k+1)+(k+2)+\cdots+N$

2 Answers2