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I am looking for cases in which $\sqrt{3n(n+1)}$ is an integer, i.e. cases in which $$ 3n(n+1)=m^2,\quad m\in\mathbb{N}. $$ I can find solutions such as $$ n=0,3,48,675,9408,131043,\dots $$ and I expect this list to be infinite. Is it? Is there a straightforward way to prove these kinds of statements?

There are many ways of reframing the problem: finding integers $n$ that are simultaneously 3 times a perfect square and 1 less than another perfect square (choosing $3n$ and $n+1$ to each be perfect squares), etc. Then I could set $n=3k^2$ and try to solve for cases in which $$ 3k^2+1=l^2 \quad\Leftrightarrow\quad 3k^2=(l+1)(l-1)\quad k,l\in\mathbb{N}. $$ It seems plausible that there are infinite solutions given the various formulas for perfect squares but none of the rabbit holes that I followed led me anywhere productive.

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    $\begingroup$ See oeis.org/A007654 $\endgroup$
    – lhf
    Jun 22, 2021 at 23:37
  • $\begingroup$ Very helpful, thank you. To be honest, even telling me to go look into Diophantine equations would have been a huge help, so this is bonus $\endgroup$ Jun 23, 2021 at 0:23

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Yeah, there's infinitely many. You can see this by considering a particular case:

if $3n$ and $n+1$ are both squares.

we can parametrize this by $n = 3k^2$.

Then we need $3k^2+1=a^2$

So we want to solve $a^2-3k^2=1$ which is a Pell equation.

the fundamental solution $(a_1,k_1)$ is $(2,1)$ and the other solutions are obtained via the recurrence $a_{n+1}=2a_n + 3k_n, k_{n+1} = a_n+2k_n$.

So we get the solutions are:

$(7,4),(26,15),(97,56), \dots$

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    $\begingroup$ And the other case is when $n$ and $3(n+1)$ are both squares, or $x^2-3y^2=3,$ for which there are no solutions $\endgroup$ Jun 23, 2021 at 0:13
  • $\begingroup$ oh ! thanks ! I thought the other case would also be easy but I was too lazy to work it out ! Thanks a lot ! $\endgroup$
    – Asinomás
    Jun 23, 2021 at 0:14
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    $\begingroup$ I had not heard of Pell equations - very helpful, especially the recurrence relation $\endgroup$ Jun 23, 2021 at 0:24
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    $\begingroup$ The recurrence comes from $a_n+k_n\sqrt3=(2+\sqrt 3)^n.$ @QuantumMechanic $\endgroup$ Jun 23, 2021 at 0:32
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From your equation, we get that $3 \mid m$, so let $m = 3j$, plus do certain manipulations, to get

$$\begin{equation}\begin{aligned} 3n(n+1) &= (3j)^2 \\ n^2 + n & = 3j^2 \\ 4n^2 + 4n & = 12j^2 \\ 4n^2 + 4n + 1 & = 12j^2 + 1 \\ (2n + 1)^2 & = 12j^2 + 1 \\ (2n + 1)^2 - 12j^2 & = 1 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Note this is a Pell's equation. For a coefficient of $12$, the fundamental solution is

$$x_1 = 7 = 2n + 1 \implies n = 3, \; \; y_1 = j = 2 \implies m = 6 \tag{2}\label{eq2A}$$

Since your initial solution of $n = 0$ makes $j = 0 \implies y_1 = 0$, it's not a positive integer so it's not included. The additional solutions section states the remaining solutions are determined from

$$2n + 1 = x_{k+1} = x_1 x_k + 12(y_1 y_k) \tag{3}\label{eq3A}$$

$$j = y_{k+1} = x_1 y_k + y_1 x_k \tag{4}\label{eq4A}$$

Since $y_1$ is even, then \eqref{eq4A} shows all $y_{k+1}$ are also even. Similarly, since $x_1$ is odd, then \eqref{eq3A} shows all $x_{k+1}$ are odd, so there's always a corresponding integer value of $n = \frac{x_{k+1}-1}{2}$.

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    $\begingroup$ Very good, thank you. My physical problem ignores the $n=0$ solution anyway so this all helps. $\endgroup$ Jun 23, 2021 at 0:25

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