Problem (Number Theory):
Kerry has a list of $n$ integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying $a_{1} \leq a_{2} \leq \ldots \leq a_{n}$. Kerry calculates the pairwise sums of all $m=\frac{1}{2} n(n-1)$ possible pairs of integers in her list and orders these pairwise sums as $s_{1} \leq s_{2} \leq \ldots \leq s_{m}$. For example, if Kerry's list consists of the three integers $1,2,4$, the three pairwise sums are $3,5,6$.


Suppose that $n=4$ and that the 6 pairwise sums are $s_{1}=8, s_{2}=104, s_{3}=106$, $s_{4}=110, s_{5}=112$, and $s_{6}=208$. Determine two possible lists $(a_{1}, a_{2}, a_{3}, a_{4})$ that Kerry could have.

Solution:
Here, the pairwise sums of the numbers $a_{1} \leq a_{2} \leq a_{3} \leq a_{4}$ are $s_{1} \leq s_{2} \leq s_{3} \leq s_{4} \leq s_{5} \leq s_{6}$. The six pairwise sums of the numbers in the list can be expressed as

$$
a_{1}+a_{2}, a_{1}+a_{3}, a_{1}+a_{4}, a_{2}+a_{3}, a_{2}+a_{4}, a_{3}+a_{4}
$$

Since $a_{1} \leq a_{2} \leq a_{3} \leq a_{4}$, then the smallest sum must be the sum of the two smallest numbers. Thus, $s_{1}=a_{1}+a_{2}$.

Similarly, the largest sum must be the sum of the two largest numbers, and so $s_{6}=a_{3}+a_{4}$. Since $a_{1} \leq a_{2} \leq a_{3} \leq a_{4}$, then the second smallest sum is $a_{1}+a_{3}$. This is because $a_{1}+a_{3}$ is no greater than each of the four sums $a_{1}+a_{4}, a_{2}+a_{3}, a_{2}+a_{4}$, and $a_{3}+a_{4}$ :

Since $a_{3} \leq a_{4}$, then $a_{1}+a_{3} \leq a_{1}+a_{4}$.

Since $a_{1} \leq a_{2}$, then $a_{1}+a_{3} \leq a_{2}+a_{3}$.

Since $a_{1} \leq a_{2}$ and $a_{3} \leq a_{4}$, then $a_{1}+a_{3} \leq a_{2}+a_{4}$.

Since $a_{1} \leq a_{4}$, then $a_{1}+a_{3} \leq a_{3}+a_{4}$.

Thus, $s_{2}=a_{1}+a_{3}$.

Using a similar argument, $s_{5}=a_{2}+a_{4}$.

So far, we have $s_{1}=a_{1}+a_{2}$ and $s_{2}=a_{1}+a_{3}$ and $s_{5}=a_{2}+a_{4}$ and $s_{6}=a_{3}+a_{4}$.

This means that $s_{3}$ and $s_{4}$ equal $a_{1}+a_{4}$ and $a_{2}+a_{3}$ in some order.

It turns out that either order is possible.

Case 1: $s_{3}=a_{1}+a_{4}$ and $s_{4}=a_{2}+a_{3}$

Here, $a_{1}+a_{2}=8$ and $a_{1}+a_{3}=104$ and $a_{2}+a_{3}=110$.

Adding these three equations gives

$$
\left(a_{1}+a_{2}\right)+\left(a_{1}+a_{3}\right)+\left(a_{2}+a_{3}\right)=8+104+110
$$

and so $2 a_{1}+2 a_{2}+2 a_{3}=222$ or $a_{1}+a_{2}+a_{3}=111$.

Since $a_{2}+a_{3}=110$, then $a_{1}=\left(a_{1}+a_{2}+a_{3}\right)-\left(a_{2}+a_{3}\right)=111-110=1$.

Since $a_{1}=1$ and $a_{1}+a_{2}=8$, then $a_{2}=7$.

Since $a_{1}=1$ and $a_{1}+a_{3}=104$, then $a_{3}=103$.

Since $a_{3}=103$ and $a_{3}+a_{4}=208$, then $a_{4}=105$.

Thus, $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(1,7,103,105)$.

Case 2: $s_{3}=a_{2}+a_{3}$ and $s_{4}=a_{1}+a_{4}$

Here, $a_{1}+a_{2}=8$ and $a_{1}+a_{3}=104$ and $a_{2}+a_{3}=106$.

Using the same process, $a_{1}+a_{2}+a_{3}=109$.

From this, we obtain $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(3,5,101,107)$.

Therefore, Kerry's two possible lists are 1,7,103, 105 and 3, 5, 101, 107.



Final answer: (1,7,103, 105), (3, 5, 101, 107)