I'm trying to write a program that would tell me whether or not a triangular number, a number of the form $\frac{(n)(n+1)}{2}$ is the sum of the squares of two other consecutive triangular numbers. It is guaranteed that the given $n$ is triangular. On oeis.org it gives a formula two calculate the nth number which satisfies the above, but no where can I find how to check whether or not a number satisfies the above.

I know this may not be the right place to post this, but I wanted a more mathematical answer to this.

  • $\begingroup$ @YvesDaoust No it is not, but if you give it an input of a triangular number, I need to check whether or not there exist two consecutive triangular numbers whose sum of squares is that $n$ $\endgroup$ – user330602 Apr 11 '16 at 20:34
  • $\begingroup$ I have updated my answer to be in line with your question $\endgroup$ – sanketalekar Apr 11 '16 at 20:37

$$8\frac{n(n+1)}2+1=\left(2n+1\right)^2.$$ Check if $8m+1$ is a perfect square. (By taking its square root.)


The question is about numbers that are the sum of two consecutive triangular numbers, i.e. which are of the form


By the above criterion, $$8m+1=(2n^2+1)^2$$ must be a perfect square and its square root $r$ must be such that $\dfrac{r-1}2$ is a perfect square.

In other words,

$$\sqrt{\frac{\sqrt{8m+1}-1}2}\in\mathbb N.$$

  • $\begingroup$ Interesting approach, I'm guessing it's a version of my approach below. How did you know to express it in terms of $(2n + 1)^2$ ? $\endgroup$ – sanketalekar Apr 11 '16 at 20:22
  • $\begingroup$ @sanketalekar: completing the square, then turning the expression to integer. $\endgroup$ – Yves Daoust Apr 11 '16 at 20:23
  • $\begingroup$ How does this work for $n=6$. It isn't supposed to work. $\endgroup$ – user330602 Apr 11 '16 at 20:23
  • $\begingroup$ @user330602: as far as I know, $49$ is a perfect square. $\endgroup$ – Yves Daoust Apr 11 '16 at 20:24
  • $\begingroup$ Yes, but you cannot find two triangular numbers whose sum of squares is 6 $\endgroup$ – user330602 Apr 11 '16 at 20:25

First Let's take this step by step to figure out the conditions for an number being triangular:

  1. For a number to be triangular, it needs to be of the form $k = \frac{n(n+1)}{2}$

  2. If we simplify this, we get the equation $2k = n^2 + n$ which becomes $n^2 + n - 2k = 0$

  3. By the Quadratic Formula, you get $n = \frac{-1 \pm \sqrt{1 + 8k}}{2} $ . Let's ignore the "-" of the $\pm$ as n is non-negative here so we get $n = \frac{-1 + \sqrt{1 + 8k}}{2} $

  4. We know n has to be a positive integer, so 1 + 8k has to be a perfect square, and the $\sqrt{1+8k} + 1$ as to be even (as it's divided by 2) so $\sqrt{1+8k}$ has to be odd. So $1 + 8k$ has to be odd, which is true regardless of k.

  5. So for a number k to be triangular, $1+8k$ has to be a perfect square, and k has to be greater than 0 (or 0 if you consider 0 triangular).

Now let's look at the conditions for a number being the square-sum of two consecutive triangular numbers. Let's consider two numbers $\frac{n(n-1)}{2}$ and $\frac{n(n+1)}{2}$

$(\frac{n(n-1)}{2})^2 +(\frac{n(n+1)}{2})^2 = \frac{n^4 -2n^3 + n^2 + n^4 + 2n^3 + n^2 }{4} = \frac{2*(n^4 + n^2)}{4} = \frac{n^2(n^2+1)}{4} $

The sum of squares of any two consecutive triangular numbers formed by n,n-1 and n+1,n is also a triangular number, with the "n" being n^2.

So for k to be triangular so that it's "n" is a perfect square

$\frac{-1 + \sqrt{1 + 8k}}{2}$ also has to be a perfect square.

So we have our conditions for k:

  1. $1+8k$ has to be a perfect square, and k has to be greater than or equal to 0.

  2. $\frac{-1 + \sqrt{1 + 8k}}{2}$ also has to be a perfect square.


We want to know if $N$ is a triangular number, in other words we want to know if there is $m$ so that $m(m+1)=2N$.

Notice that if $m=\sqrt{2n}$ we get $m(m+1)=2n+\sqrt{2n}>n$.

Notice if $m=\sqrt{2n}-1$ we get $m(m+1)=2n-\sqrt{2n}<n$

So we only have to try with numbers in the range $(\sqrt{2n}-1,\sqrt{2n})$, so we need only check if $\lfloor\sqrt{2n}\rfloor(\lfloor\sqrt{2n}\rfloor+1)=2n$


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