A question involving integers and adjacent and diagonal sums. I have the group of integers $1 - 10$, and I'm trying to arrange them in a circle in the $xy$-plane such that the sum of any two adjacent integers equals the sum of their diagonally opposite integers. 

I am trying to figure out a way to express this using math. I've tried looking at the pattern and then reducing the set to integers from $1 - 8$ and $1-6$, but the pattern from integers $1 -  10$ doesn't work for it. I am new to MSE.
 A: I shall label the numbers in the circle as follows:
\begin{matrix}
a_0 & a_1 & a_2 & a_3 \\
a_9 &     &     & a_4 \\
a_8 & a_7 & a_6 & a_5
\end{matrix}
Consider, for example, the pair $a_0$ and $a_1$.
Their sum must be the same as the numbers opposite them, i.e.
\begin{align}
a_0+a_1 & = a_5 + a_6 \\
\implies a_0-a_5 & = a_6 - a_1
\end{align}
And similarly, if we look at the pair $a_1$ and $a_2$, we find that
\begin{align}
a_1+a_2 & = a_6 + a_7 \\
\implies a_6-a_1 & = a_2 - a_7
\end{align}
Continuing around the circle, we obtain
$$a_0-a_5 = a_6-a_1 = a_2-a_7 = a_8 - a_3 = a_4 - a_9 \qquad \qquad (*)$$
i.e. The five pairs of opposite numbers $\{a_0,a_5\}$, $\{a_1,a_6\}$, $\{a_2,a_7\}$, $\{a_3,a_8\}$, $\{a_4,a_9\}$ must have the same differences between them. It is clear that the common difference must be $5$, and that the five pairs are precisely $\{1,6\}$, $\{2,7\}$, $\{3,8\}$, $\{4,9\}$, $\{5,10\}$ in some order.
This is pretty much it, as the solution is not unique. For example, you can have
\begin{matrix}
10 & 1 & 7 & 3 \\
4 &     &     & 9 \\
8 & 2 & 6 & 5
\end{matrix}
Or maybe
\begin{matrix}
9 & 1 & 8 & 2 \\
5 &     &     & 10 \\
7 & 3 & 6 & 4
\end{matrix}
As long as opposite pairs have a difference of $5$, and you have the signs the right way round (refer to the $*$ equation above).
