# Are there an infinite number of integers $n$ such that $(3n)(n+1)$ is a perfect square?

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.

• – lhf
Jun 22, 2021 at 23:37
• 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 Jun 23, 2021 at 0:23

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:

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

• 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 Jun 23, 2021 at 0:13
• oh ! thanks ! I thought the other case would also be easy but I was too lazy to work it out ! Thanks a lot ! Jun 23, 2021 at 0:14
• I had not heard of Pell equations - very helpful, especially the recurrence relation Jun 23, 2021 at 0:24
• The recurrence comes from $a_n+k_n\sqrt3=(2+\sqrt 3)^n.$ @QuantumMechanic Jun 23, 2021 at 0:32

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 $$n = \frac{x_{k+1}-1}{2}$$.

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