Close Encounter of the Triangular Kind

[This is not a complete solution. Indeed, in the following it is used a series identity that I found (the one coloured in $\color{#3D99F6} \text{blue}$ ) of which I don't know a proof nor I have a link to one. All hints are welcome.]

A positive integer $T$ is a triangular number if and only if $8T+1$ is a perfect square. So, in the problem we have $48k+1=x^2$ , for $x \in \mathbb{Z}$ . Hence, $x^2 \equiv 1 \; \text{mod}\, 48$ and then $k = \frac{x^2-1}{48}$ .

Now, each $0 \le b \le 47$ such that $b^2 \equiv_{48} 1$ generates a set $K_b=\Bigl \{ \dfrac{(b+48h)^2-1}{48}, \, h \in \mathbb{Z} \Bigr \}$ of solutions for $k$ . Since, moreover, every solution is contained in a set of this form, we have that $K$ is the union of all $K_b$ . Let also $S_b$ denote the sum of all the reciprocals of elements of $K_b$ .

For two distinct $b_1$ , $b_2$ of this kind, $K_{b_1}$ and $K_{b_2}$ are not disjoint if and only if $b_2=48-b_1$ . Indeed, if there is a common element, then exist $h_1,h_2$ for which $(b_1+48h_1)^2=(b_2+48h_2)^2$ , and so $\pm (b_1+48h_1)=b_2+48h_2$ . Thus $b_1 \equiv_{48} b_2$ $\lor$ $-b_1 \equiv_{48} b_2$ . But since $b_1 \ne b_2$ and they lie in the interval above, $b_2=48-b_1$ . Conversely, given $b_2=48-b_1$ , if an integer $h$ determines by the respective formula an element of $K_{b_1}$ or $K_{b_2}$ , then we can plug $-(h+1)$ in the other formula to see that the element also belongs to the other set. Moreover, notice that $b_1$ and $b_2$ actually generate the same set of solutions, therefore if $K_{b_1}$ and $K_{b_2}$ are not disjoint, they are equal.

Hence, we conclude that all integers $b$ between $0$ and $24$ such that $b^2 \equiv_{48} 1$ provide pair-wise disjoint sets $K_b$ and that their union equals $K$ . The possible values for $b$ are $1,7,17,23$ . Thus

$\displaystyle S = \sum_{k \in K} \frac{1}{k} = S_1+S_7+S_{17}+S_{23}$

Knowing that $\displaystyle \sum_{k= 1}^{\infty} \frac{1}{k^2-z^2} = \frac{1}{2z} \Bigl ( \frac{1}{z}-\pi\cot(\pi z) \Bigr)$ , we can evaluate

$\displaystyle S_1 = 48 \sum_{h=1}^{\infty} \frac{1}{(1+48h)^2-1} + \frac{1}{(1-48h)^2-1} = \frac{1}{24} \sum_{h=1}^{\infty} \frac{1}{h^2- \frac{1}{24^2}} = 12- \frac{1}{2} \pi \cot\frac{\pi}{24}$

$\,$

While for $b \ne 1$ , $\color{#3D99F6} \displaystyle S_b = 48 \sum_{h=-\infty}^{\infty} \frac{1}{(b+48h)^2-1} = \frac{\pi}{2} \Bigl (\cot\frac{\pi b-\pi}{48}- \cot\frac{\pi b+\pi}{48} \Bigr)$

$\,$

Hence

$\displaystyle S = 12+ \frac{\pi}{2} \Bigl ( -\cot \frac{\pi}{24}+\cot \frac{\pi}{8}- \cot \frac{\pi}{6} + \cot\frac{\pi}{3} - \cot\frac{3}{8} \pi + \cot\frac{11}{24}\pi - \cot \frac{\pi}{2 }\Bigr)$

However, if $\alpha + \beta = \dfrac{\pi}{2}$ , with $\alpha,\beta \ne k\pi$ , then $\cot \alpha \cot \beta = 1$ and so $\cot \alpha = \tan \beta$ . For instance, $\cot\dfrac{11}{24}\pi = \tan \dfrac{\pi}{24}$ . So

$\displaystyle S = 12+ \frac{\pi}{2} \Bigl ( \tan\frac{\pi}{24} - \tan\frac{\pi}{8} - \tan \frac{\pi}{3} - \Bigl( \cot \frac{\pi}{24}-\cot \frac{\pi}{8} - \cot\frac{\pi}{3}\Bigr) \Bigr)$

And $(A-B)^2+C^2 = (8-3)^2+12^2 = \boxed{169}$

If $6k$ is a triangular number, it may be written as $6k = \tfrac12n(n\pm 1)$ . Double this to find the equation $12k = n(n\pm 1).$ Since both sides of this equation are multiples of 12, we have three possibilities

• $n$ is a multiple of 12, say $n = 12c$ . Then $k = 12c^2 \pm c$ for $c = 1, 2, \dots$ are all distinct solutions. Thus the sum contains the terms $\sum_{c=1}^\infty \frac 1{12c^2 \pm c} = \sum_{c=1}^\infty \left(\frac 1{c\pm \tfrac1{12}} - \frac 1c\right).$

• $n$ is a multiple of 6 but not of 12, say $n = 6c$ for odd $c$ . Then $k = 3c^2 + \tfrac12c$ . But since $c$ is odd, this cannot be an integer.

• $n$ is a multiple of 4 but not of 12, say $n = 4a$ . Then $n\pm 1 = 4a\pm 1$ must be a multiple of 3. But this implies that $a = 3c \mp 1$ . It is now easy to show that $k = 12c^2 \pm 7c + 1$ . For $c = 1, 2, \dots$ this gives distinct solutions; for $c = 0$ we find the additional solution $k = 1$ . Thus the sum contains the terms $1 + \sum_{c=1}^\infty \frac 1{12c^2 \pm 7c + 1} = 1 + \sum_{c=1}^\infty \left(\frac 1{c\pm \tfrac1{3}} - \frac 1{c\pm \frac 14}\right).$

Therefore the required sum is $\sum_{k \in K} \frac 1k = 1 + \sum_{c=1}^\infty \left(\frac 1{c+ \tfrac1{12}} + \frac 1{c- \tfrac1{12}} - \frac 2c + \frac 1{c+ \tfrac1{3}} + \frac 1{c- \tfrac1{3}}- \frac 1{c+ \frac 14}- \frac 1{c- \frac 14}\right).$ Add some magic to find that this sum is actually equal to $\sum_{k \in K} \frac 1k = 12 - \pi \left(\frac 4{\sqrt 3} + 1\right),$ but for unknown reason we must express the term in brackets using a cotangent. Now $\cot \pi/3 = 1/\sqrt 3$ , so that $\sum_{k \in K} \frac 1k = 12 - \pi \left(4\cot\frac\pi 3 + 1\right).$ We see that $A = 12$ , $B = 4$ , $C = 3$ , and $A^2 + B^2 + C^2 = 3^2 + 4^2 + 12^2 = \boxed{169}$ .

Note: Since this solution was posted, the problem has been edited. The answer is still 169.