How far apart can we be?

Denote $A_i$ as a vector. Then the sum in question is : $\sum_{1 \le i \le j \le 15} (A_i - A_j) \cdot (A_i - A_j) = \sum_{1 \le i \le j \le 15} A_i^2 - 2A_i \cdot A_j + A_j^2.$

It is clear this sum is then:

$\begin{array}{ll} & \quad \sum_{i=1}^{15} 14A_i^2 - \sum_{1 \le i \neq j \le 15} A_i \cdot A_j \\ & = \sum_{i=1}^{15} 15A_i^2 - \sum_{1 \le i,j \le 15} A_i \cdot A_j \\ & = \sum_{i=1}^{15} 15A_i^2 - (A_1 + A_2 + ... + A_{15}) \cdot (A_1 + A_2 + ... + A_{15}) \\ & \le \sum_{i=1}^{15} 15A_i^2 \\ & \le \sum_{i=1}^{15} 15 = 225 \end{array}$

Equality occurs by having an equilateral triangle on the sphere and then 5 points at each vertex, so we are done.

[Note: Equality occurs when $|A_i |=1$ and $\sum A_i=0$ . There are many equality cases, like using any 5 equilateral triangles with circumcenter at the origin and radius 1.]

We make the claim that if we place $n>1$ points on the unit sphere, then the sum of the squares of the pairwise differences is less than or equal to $n^2$ .

To prove this, let the points $A_1, A_2, \ldots, A_{n}$ be placed at the end of the unit vectors $\mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_{n}}$ , respectively. This simplifies our problem into evaluating the maximum value of $\displaystyle\sum_{1\le i<j\le n} |\mathbf{v_i}-\mathbf{v_j}|^2$

We note that $|\mathbf{v_i}-\mathbf{v_j}|^2=|\mathbf{v_i}|^2-2\mathbf{v_i}\cdot \mathbf{v_j} + |\mathbf{v_j}|^2$ . In order to get rid of the $2\mathbf{v_i}\cdot\mathbf{v_j}$ terms, we can add by $|\mathbf{v_1}+\mathbf{v_2}+\ldots+\mathbf{v_n}|^2$ to get the equation

$|\mathbf{v_1}+\mathbf{v_2}+\ldots+\mathbf{v_n}|^2+\displaystyle\sum_{1\le i<j\le n} |\mathbf{v_i}-\mathbf{v_j}|^2 =$ $n(|\mathbf{v_1}|^2+|\mathbf{v_2}|^2+\ldots+|\mathbf{v_n}|^2)=n(1(n))=n^2\implies$ $\displaystyle\sum_{1\le i<j\le n} |\mathbf{v_i}-\mathbf{v_j}|^2 = n^2-|\mathbf{v_1}+\mathbf{v_2}+\ldots+\mathbf{v_n}|^2\implies$

$\displaystyle\sum_{1\le i<j\le n} |\mathbf{v_i}-\mathbf{v_j}|^2 \le n^2$

Thus, the pairwise sum is maximized when $|\mathbf{v_1}+\mathbf{v_2}+\ldots+\mathbf{v_n}|^2$ is minimized, and this occurs when $\mathbf{v_1}+\mathbf{v_2}+\ldots+\mathbf{v_n}=0$ , so the maximum value is $15^2=\boxed{225}$ . This is easily accomplished, placing $5$ of $A_i$ at each vertex of a equilateral triangle inscribed in a great circle of the unit sphere.

Let the vectors of each of the 15 points be ${u_i}$ corresponding to the points ${A_i}$ . For a vector $v$ , let $v^2 = |v|^2 = v \cdot v$ for simplicity.

Thus, we have $|A_i A_j|^2 = |u_i - u_j|^2 = (u_i - u_j) \cdot (u_i - u_j) = u_i^2 - 2u_i \cdot u_j + u_j^2$ .

The summation becomes $14 \displaystyle \sum_{i=1}^{15} u_i^2 - 2 \displaystyle \sum_{1 \leq i<j \leq 15} u_i \cdot u_j$ . The terms motivate the expansion of $(u_1 + \ldots + u_{15})^2 = \displaystyle \sum_{i=1}^{15} u_i^2 + 2 \displaystyle \sum_{1 \leq i<j \leq 15} u_i \cdot u_j$ .

Hence, the original sum may be rearranged to $15 \displaystyle \sum_{i=1}^{15} u_i^2 - (u_1 + \ldots + u_{15})^2$ . The maximum value occurs when $u_i^2 = 1$ for $1 \leq i \leq 15$ and $(u_1 + \ldots + u_{15})^2=0$ . This is easily done by selecting the roots of unity of order 15 as vectors. Hence, the sum is maximized at $15*15=225$ .

Let the $n$ points $A_i(x_i,y_i,z_i), \forall i=1,2,...,n$ be on or inside the unit sphere $x^2+y^2+z^2=1$ . The square of the distance between the pair of points $(i,j)$ such that $i<j$ is:

$(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2$

$=x_i^2+y_i^2+z_i^2+x_j^2+y_j^2+z_j^2-2(x_ix_j+y_iy_j+z_iz_j)$

$\leq 2-2(x_ix_j+y_iy_j+z_iz_j)$ .

Note that equality will occur if all the points are on the surface of the sphere (i.e., $x_i^2+y_i^2+z_i^2=1, \forall i$ ).

Since there are $\frac{n(n-1)}{2}$ such pairs of $(i,j)$ ,

$m:=\displaystyle\sum_{1\leq i<j\leq n} \left| A_i A_j \right|^2 \leq n(n-1)-2 \displaystyle\sum_{1\leq i<j\leq n} (x_ix_j+y_iy_j+z_iz_j)$

Now, $\left(\displaystyle\sum_{i=1}^n x_i\right)^2=\displaystyle\sum_{i=1}^n x_i^2+2\displaystyle\sum_{1\leq i<j\leq n} x_ix_j$ . Thus,

$2\displaystyle\sum_{1\leq i<j\leq n} (x_ix_j+y_iy_j+z_iz_j)=X^2+Y^2+Z^2-\displaystyle\sum_{i=1}^n (x_i^2+y_i^2+z_i^2)$

where $X=\left(\displaystyle\sum_{i=1}^n x_i\right)^2, Y=\left(\displaystyle\sum_{i=1}^n y_i\right)^2$ , and $Z=\left(\displaystyle\sum_{i=1}^n z_i\right)^2$ .

Therefore,

$m\leq n^2-n-X^2-Y^2-Z^2+\displaystyle\sum_{i=1}^n (x_i^2+y_i^2+z_i^2)\leq$ $n^2-n-X^2-Y^2-Z^2+n=n^2-X^2-Y^2-Z^2$ .

Again, equality will occur if all points are on the surface of the sphere.

Thus, $m\leq n^2$ . In this case, $n=15$ and $m\leq 225$ . Equality will occur if $X^2=Y^2=Z^2=0$ , which is the case if the centroid of the points is the origin.

Note that $|A_i A_i|^2 =0$ and doesn't affect the sum. Let the points be in vector form. Let $\vec{X} = \sum_{i=1}^{15} \vec{A_i}$ . For each $k$ , we have

$\begin{aligned} \sum_{i=1}^{15} |A_i A_k|^2 &= \sum_{i=1}^{15} (\vec{A_i} - \vec{A_k} ) \cdot (\vec{A_i} - \vec{A_k}) \\ &= \sum_{i=1}^{15} \left(\vec{A_i} \cdot \vec{A_i} - 2 \vec{A_i}\cdot \vec{A_k} + \vec{A_k} \cdot \vec{A_k}\right) \\ &\leq 15 + 15 - 2 \vec{A_k} \sum_{i=1}^{15} \vec{A_i} \\ &= 30 - 2 \vec{A_k} \cdot \vec{X} \\ \end{aligned}$

Hence $\sum_{i < j} |A_i A_j| ^2 \leq \frac {1}{2} \sum_{i=1}^{15} \sum_{j=1}^{15} |A_i A_j|^2 = \sum_{i=1}^{15} \left(15 - \vec{A_i} \cdot \vec{X}\right) = 225 - \vec{X} \cdot \vec{X} \leq 225$ .

Equality holds if and only if $\vec{X} = 0$ and $|A_i|=1 \, \forall i$ , which occurs in a regular 15-gon with circumradius 1 (and there are numerous more examples). Hence, the maximal value is 225.