Determine the integers $m$ such that $1+2+3+…m=$ a perfect square

Question

I went like this

$ \frac{m(m+1)}{2} = k^2 ; m^2+m = 2k^2$

From here I noticed that m is even that is $m =2q$ Substituting it will give

$ 2q^2+q = k^2$

From here I got no where..

Then I thought perfect square can be expressed as sum of consecutive odd number.I tried that way it too ended similarly. Any ideas for progress??

Answer 1

So, $4q^2+4q+1=2k^2+1$, that is $r^2-2k^2=1$ where $r=2q+1$. This is a case of Pell's equation.

Answer 2

The question can be said as there is a number which is the $m$th triangular number and also $k$th square number. Actually, there is a number called square triangular number. There is a sequence of finding the $m$ and $k$.

Answer 3

(Since only half of the solutions were addressed so far, or none sufficiently explicitly, let me provide all of them).

We have to solve

$$ m\cdot(m+1)\ =\ 2\cdot k^2 $$

i.e. $\ 2\cdot k^2\ =\ L\cdot M\ $ where $\ \{L\ M\}=\{m\,\ m+1\}\ $ is a product of two consecutive natural numbers. Then one of them, say $\ M,\ $ has to be odd, and consequently, the other one, $\ L,\ $ has to be even; it follows that $\ L=2\cdot\lambda^2\ $ and $\ M=\mu^2\ $ for certain natural numbers $\ \lambda\ \mu.\ $ In effect, we obtain equation

$$ |L-M| = 1 $$ i.e. $$ |2\cdot\lambda^2\ -\ \mu^2|\ =\ 1 $$

It's well known (and not too hard to prove) that all solutions form the following sequence:

$$ \lambda_0:=\mu_0:=1 $$ and $$ \forall_{n\in\mathbb N}\quad (\,\lambda_n:=\lambda_{n-1}+\mu_{n-1}\,\ \mbox{and} \ \ \mu_n:=\lambda_n+\lambda_{n-1}\,) $$ Thus, finally:

$$ m_s :=\ \mu_s^2\qquad \forall\ s\ \mbox{odd} $$ and $$ m_t :=\ 2\cdot\lambda_t^2\qquad \forall\ t\ \mbox{even} $$

and, of course,

$$ \forall_{r\in\mathbb N}\quad k_r\ =\ \lambda_r\cdot\mu_r $$

provides all solutions $\ (m\ k)\ :=\ (m_r\ k_r),\ $ where $\ r\in\mathbb N$.