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If $$N = \frac{{n_1}(n_1+1)}{2}\cdot{d_1},$$ is it possible to have $$N = \frac{{n_2}(n_2+1)}{2}\cdot{d_2}?$$

Here, $d_i, n_i \in \mathbb{N}$, $d_i > 1$ and $n_1 \neq n_2$.

That is,

Question 1

If a number can be represented as a nontrivial multiple of a triangular number, can it then be represented as a nontrivial multiple of another triangular number?

Added January 27 2017

and

Question 2

If such a number can be so represented (as in Question 1), must it necessarily be (a nontrivial multiple of or) the least common multiple of the triangular numbers $$T(n_1)=\dfrac{{n_1}(n_1 + 1)}{2}$$ and $$T(n_2)=\dfrac{{n_2}(n_2 + 1)}{2}?$$

Thanks to users Alex Macedo, dxiv, and Stahl for their initial comments!

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    I don't know if I understand the question. Maybe $N$ is the product of two distinct triangular numbers. – 2017-01-27
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    @AlexMacedo, that qualifies as an answer! However, I would be more interested in the case when it isn't. – 2017-01-27
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    @JoseArnaldoBebitaDris $3,6,36$ are all triangular numbers. – 2017-01-27
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    More generally, $N$ could be the least common multiple of two distinct triangular numbers. – 2017-01-27
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    $N_{n,d} = \frac{(n^2 - 1)((n^2-1) + 1)}{2}\cdot d = \frac{(n + 1)n}{2}\cdot\left((n - 1)n d\right)$ is another class of examples for $n > 1$, $d > 1$ ($n_1 = n^2 - 1$, $n_2 = n$). – 2017-01-27
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    These are all very good comments and directly answer my original question. I invite you to post an answer, and I will accept the best one. – 2017-01-27
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    @JoseArnaldoBebitaDris regarding your second question, I just want to point out that you ask for $N$ to be a multiple of both $T(n_1)$ and $T(n_2)$, for some $n_1$ and $n_2$. This is the same as asking for $N$ to be a multiple of $\operatorname{lcm}(T(n_1), T(n_2))$. – 2017-01-28

1 Answers 1

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Let $\dfrac{n_1(n_1+1)}{n_2(n_2+1)} = \dfrac{d_2}{d_1}$

It follows that $\dfrac{{n_1}(n_1+1)}{2}\cdot{d_1} = \dfrac{{n_2}(n_2+1)}{2}\cdot{d_2}$

Example $T(6) = 21$ and $T(8)=36$. So $\dfrac{T(6)}{T(8)} = \dfrac{7}{12}$

Hence $\dfrac{{6}(6+1)}{2}\cdot{12} = \dfrac{{8}(8+1)}{2}\cdot{7}$

If $n_1$ and $d_1$ are given {It would have been nice if you were a bit more explicit}, then you want to find $n$ and $d$ that solve

$\dfrac{{n}(n+1)}{2}\cdot{d} = \dfrac{n_1(n_1+1)}{2}\cdot{d_1}$

Which can be expressed as

$n^2 + n - \dfrac{n_1(n_1+1) d_1}{d} = 0$

and can be solved by inspection.

Example

\begin{align} \dfrac{4(4+1)}{2}\cdot{3} &= \dfrac{{n}(n+1)}{2}\cdot{d} \\ n^2 + n - \dfrac{60}{d} &= 0 \\ n &= \dfrac{-1 + \sqrt{1+\dfrac{240}{d}}}{2} \\ \end{align}

The possible integer values of $\dfrac{240}{d}$ are $$\{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240\}$$

We note that $1+\{3, 8, 15, 24, 48, 80, 120\}$ are perfect squares.

Which leads to $d \in \{80, 30, 16, 10, 5, 3, 2\}$

and $n \in \left\{\dfrac 12, 1, \dfrac 32, 2, 3, 4, 5\right\}$.

Removing the fractions, we end up with $(n,d) = \{(1,30), (2,10), (3,5), (4,3), (5,2) \}$

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    Thank you for your comprehensive answer, Steven! This is exactly what I was expecting... =) – 2017-01-28