1
$\begingroup$

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??

$\endgroup$
  • $\begingroup$ The observation about $\ m\ $ being even is not obvious. $\endgroup$ – Wlod AA Sep 15 '19 at 7:49
  • $\begingroup$ It's even FALSE: $\ m=k=1\ $ gives a counter-example. $\endgroup$ – Wlod AA Sep 15 '19 at 7:51
  • $\begingroup$ Another quick solution is $\ m=8\ $ with $\ k=6.$ $\endgroup$ – Wlod AA Sep 15 '19 at 7:55
  • 1
    $\begingroup$ Oh thanks i forget m could be 1 mod 2 $\endgroup$ – Mathematical Curiosity Sep 15 '19 at 7:56
  • $\begingroup$ Did you do it by trial and error?? $\endgroup$ – Mathematical Curiosity Sep 15 '19 at 7:57
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Can any one tell me how to write in lines?? When I write it automatically merges all lines $\endgroup$ – Mathematical Curiosity Sep 15 '19 at 7:30
  • $\begingroup$ Try pressing Enter to leave a line gap between places you want to have separate lines. $\endgroup$ – Minus One-Twelfth Sep 15 '19 at 7:38
1
$\begingroup$

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 $k$.

$\endgroup$
1
$\begingroup$

(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 $\ r\in\mathbb N$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.