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Context:

A triangular number is a number of the form $n(n+1)/2$ where $n\in \mathbb N$.

Question:

After application of two theorems in my textbook (the first one being that "a number $n$ is triangular if and only if $\exists N\in \mathbb N,\text{ }8n+1 = N^2$" the second one being "if a number $k$ is odd then $\exists p \in \mathbb N, \text{ } k^2 = 8p + 1$") the following result came out:

To every odd number $k$, we can assign a triangular number $n$, namely: $\frac{k^2-1}{8}$.

My question is two-fold:

  • does the converse hold? That is, can we assign to every triangular number an odd number?
  • if the answer of question $1$ is yes, is there by any chance, a way to assign to a every even number a special number in direct analogy with the way it happens for odd numbers?

Thanks

niobium
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    What happens when you solve $n = \frac{k^2 - 1}{8}$ for $k$? – Travis Willse Sep 03 '23 at 19:57
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    The second question is easy since there is a simple $1-1$ correspondence between the odd and the even numbers , namely $f(n)=n+1$ mapping the odds to the evens and $g(n)=n-1$ for mapping the evens back to the odds. For the first , you only need that the triangle numbers form a countable set. Hence , there must be a $1-1$ correspondence between the odd numbers and the triangle numbers. – Peter Sep 03 '23 at 19:59
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    Maybe it is worth mentioning that every non-negative integer is the sum of at most three triangle numbers. – Peter Sep 03 '23 at 20:11
  • @Peter thanks for the mapping between the even and the odds, but I don't fully get what countability has to do with the triangular numbers – niobium Sep 03 '23 at 20:25
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    Niobium, what Peter is getting at is that countability ensures that such assignments are abundant and not particularly special, if you don't put any extra constraints on them. For instance, consider the following silly mapping from triangular numbers to odd numbers: given $T_n$, choose the odd number $2T_n+1$. Clearly this is odd, and no odd number is hit by two different triangulars, so technically it does the job. It also has nothing to do with the nice arithmetic correspondence in the other direction; when asking for existence, we should set up our requirements to avoid such trivialities. – Eric Nathan Stucky Sep 03 '23 at 23:04

1 Answers1

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Note that $$\frac {k^2-1}8=\frac 12 \cdot \frac {k-1}2 \cdot \frac {k+1}2$$ and setting $k=2n+1$ we obtain $$\frac {k^2-1}8=\frac {n(n+1)}2$$

So if we have a triangular number we can assign the odd number $k=2n+1$ and the arithmetic all works.

It is artificial to do anything with even numbers in the same way, because essentially you would be looking for even numbers to map to somewhere in the gaps between triangular numbers. You can, of course simply plug $k=2n$ into the formula you have, but that will not give you an integer. If you have to change the formula to get what you want, that is an indication that the generalisation may not be fruitful.

Mark Bennet
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  • We can simply assign an even number by adding one to the assigned odd number. I cannot see a problem. The formula to recover the triangle number has only slightly to be modified to do the job. – Peter Sep 04 '23 at 14:48