Recently I tried to solve a diophantine equation

$n(n+1)=4m(m+1)$ with $n,m\in\mathbb{Z}$

which resulted from an other equation. But how can one show, that there are no non-trivial solutions.

Obvsiously there are four solutions. But is there an elemantary way to show, that there are no more?

Thanks in advance.


Multiplying by $4$ and rewriting, you get:

$$(2n+1)^2-1 = 4\left[(2m+1)^2-1\right]$$

Rewriting, you get: $$(4m+2)^2-(2n+1)^2 = 3$$

This means $$(4m+2n+3)(4m-2n+1)=3.$$

  • 1
    $\begingroup$ Elegant solution! $\endgroup$ – Cornman Jul 27 '18 at 16:02
  • $\begingroup$ It's a fairly standard approach to solving diophantine equations if the form $p(m)=q(m)$ when $p,q$ are quadratic. You usually get something "Pell-like." $\endgroup$ – Thomas Andrews Jul 27 '18 at 16:07
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    $\begingroup$ I personally like better stopping at the second line, noting that if the difference between two squares is $3$, then the two squares must be $4$ and $1$. That way it's a bit more straight-forward to isolate $m$ and $n$ from one another. $\endgroup$ – Arthur Jul 27 '18 at 16:07

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