Integer solutions of k(k+1)/2=m^2

Asked Mar 29 '20 at 08:16

Active Mar 29 '20 at 08:18

Viewed 34 times

When I was solving the problem,

"Prove that there are infinity integer solutions for $x^2-2y^2=1$"

I did this kind of thing:

Since x is odd and y is even, let $x=2k+1$ and $y=2m$ then we get $k(k+1)/2=m^2$.

Since there are infinity amount of integers for $x^2-2y^2=1$,

we result that k(k+1)/2=m^2 has infinity amount of integer solutions.

I think we can prove that k(k+1)/2=m^2 has infinity amount of integer solutions in some way.

(Sum of 1~k) = (m^2) ---> this looks kind of cool

My question is :

"Prove that $k(k+1)/2=m^2$ has infinitely many integer solutions. (without using $x^2-2y^2=1$)"

edited Mar 29 '20 at 08:18

Mostafa Ayaz

asked Mar 29 '20 at 08:16

Sanghwa Lee

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Mar 29 '20 at 08:22
"Let $x=2k+1$ and $y=2m$ then we get $k(k+1)/2=m^2$". So $k^2+k=2m^2$. This is just rewriting Pell. – Dietrich Burde Mar 29 '20 at 08:57
@DietrichBurde Thanks for the comment. Btw I'm curious if (sum of 1~k)=(m^2) has some kind of algebraic or combinational meaning. – Sanghwa Lee Mar 29 '20 at 12:54
The sum of consecutive integers has a combinatorial meaning, see here, but that this is again a square has an algebraic meaning - again given by Pell's equation and its algebraic solution. So we are back again. – Dietrich Burde Mar 29 '20 at 13:07

0 Answers0