0

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

The solution says "it's easy to check that $(k,N)$ solution $\implies$ $(3k+2N+2,4k+3N+3)$ solution too". How do we verify that? So far I found $$1+2+\cdots+k = \dfrac{k(k+1)}{2}= (k+1)+(k+2)+\cdots+N = \dfrac{(N-k)(N+k+1)}{2}$$ and $$1+2+\cdots+(3k+2N+2) = \dfrac{(3k+2N+3)(3k+2N+2)}{2}= (3k+2N+3)+(3k+2N+4)+\cdots+(4k+3N+3) =\dfrac{(7k+5N+6)(N+k+1)}{2}.$$ Is there an easier way?

Caleb Stanford
  • 45,895
user19405892
  • 15,592

2 Answers2

1

We can check that $(k,N)$ is a solution $\Leftrightarrow (2N+1)^{2}-2(2k+1)^{2}=-1$, which is called Pell's equation. If we consider multiplicative norm $N(a+b\sqrt{2})=a^{2}-2b^{2}$ of a ring $\mathbb{Z}[\sqrt{2}]$, we can check that if $(x,y)$ is a solution then also $(3x-4y, -2x+3y)$ is. ($N((3x-4y)+(-2x+3y)\sqrt{2})=N((3-2\sqrt{2})(x+y\sqrt{2}))=N(3-2\sqrt{2})N(x+y\sqrt{2})=N(x+y\sqrt{2})$.) This is the way how such map $(k,N)\to (3k+2N+2, 4k+3N+3)$ was born. In fact, $(2N+1)+(2k+1)\sqrt{2}=x_{n}+y_{n}\sqrt{2}=(\sqrt{2}-1)^{2n+1}$ gives the complete set of the solutions. (You can find proof in any number theory textbook that explains about Pell's equation.)

Seewoo Lee
  • 15,137
  • 2
  • 19
  • 49
0

I don't see an easier way, but you are almost there. Clearing the $2$ in the denominator, you can say $$(3k+2N+3)(3k+2N+2)=9k^2+12kN+4N^2+15k+10N+6\\ (7k+5N+6)(N+k+1)=7k^2+12kN+5N^2+13k+11N+6$$The difference is $2k^2-N^2+2k-N=2k(k+1)-N(N+1)$, which is zero by your first equality.

Ross Millikan
  • 374,822
  • How do you think the solution knew to think of that pair? – user19405892 Jul 19 '16 at 03:55
  • Also, how is it zero by my first equality? – user19405892 Jul 19 '16 at 03:58
  • I don't know, but it looks like it could have come from the continued fraction and some exploration. If you start from $N(N+1)=2k(k+1)$, which is your first equality, you want $N \approx k\sqrt 2$ so you look at the convergents to $\sqrt 2$. You could look at Pell's equation to get started. – Ross Millikan Jul 19 '16 at 04:01