10. Arrange the positive integers that can be expressed as the difference of squares of two positive integers in ascending order: $a_{1}, a_{2}, \cdots$, $a_{n}, \cdots$ (for example, $a_{1}=2^{2}-1^{2}=3, a_{2}=3^{2}-2^{2}=5, a_{3}$ $\left.=4^{2}-3^{2}=7, a_{4}=3^{2}-1^{2}=8, \cdots\right)$. Then, the value of $a_{1}+$ $a_{2}+\cdots+a_{10}$ is $\qquad$
Two, (20 points) As shown in Figure 5, given two concentric circles with radii of $1$ and $2$, there is a square $A B C D$, where points $A$ and $D$ lie on the circumference of the circle with radius $2$, and points $B$ and $C$ lie on the circumference of the circle with radius $1$. Find the area of this square.
10. 6999.

First, it is evident that even numbers that are not multiples of 4 cannot be the difference of squares of two integers. It is known that \(a_{1}=3, a_{2}=5, a_{3}=7\).
When \(k \geqslant 2\), we have
\[4 k=(k+1)^{2}-(k-1)^{2},\]
\[4 k+1=(2 k+1)^{2}-(2 k)^{2},\]
\[4 k+3=(2 k+2)^{2}-(2 k+1)^{2},\]

and \(4 k+(4 k+1)+(4 k+3)=12 k+4\).
Thus, \(a_{4}+a_{5}+a_{6}=12 \times 2+4\),
\[a_{7}+a_{8}+a_{9}=12 \times 3+4,\]
\[
\begin{array}{l}
a_{97}+a_{98}+a_{99}=12 \times 33+4, \\
a_{100}=4 \times 34 .
\end{array}
\]

Therefore, \(a_{1}+a_{2}+\cdots+a_{100}\)
\[
\begin{array}{l}
=3+5+7+12(2+3+\cdots+33)+4 \times 32+4 \times 34 \\
=6999 .
\end{array}
\]

Second, as shown in Figure 7, connect
\(O C, O D\). Draw \(O E \perp A D\), with the foot of the perpendicular at \(E\), intersecting \(B C\) at point \(F\). Let the side length of the square be \(2 x\). Clearly, \(E\) and \(F\) are the midpoints of \(A D\) and \(B C\), respectively. By the Pythagorean theorem, we have \(O E=\sqrt{4-x^{2}}\). Since
\[
\begin{array}{l}
O F=|O E-E F| \\
=\left|\sqrt{4-x^{2}}-2 x\right|, \\
C F^{2}+O F^{2}=O C^{2},
\end{array}
\]

we have \(x^{2}+\left(\sqrt{4-x^{2}}-2 x\right)^{2}=1\), which simplifies to \(4 x^{2}+3=4 x \sqrt{4-x^{2}}\).
Squaring both sides and simplifying, we get \(32 x^{4}-40 x^{2}+9=0\).
Solving this, we find \(x^{2}=\frac{5 \pm \sqrt{7}}{8}\).
Thus, the area of square \(A B C D\) is \(4 x^{2}=\frac{5 \pm \sqrt{7}}{2}\).