A sequence that starts with a positive number has the property that each of the following terms is the perimeter of the square with area equal to the preceding term. If the first three terms form an arithmetic sequence, what are the possible values for the first term of the sequence? (Having a common difference of 0 is allowed.)
Let the first term of the sequence be $a$. If $a$ is the area of a square, then the side length of that square must be $\sqrt{a}$, so the second term must be $4 \sqrt{a}$. Similarly, the third term must be $4 \sqrt{4 \sqrt{a}}=8 \sqrt[4]{a}$. If these terms form an arithmetic sequence, then they have a common difference so that

$$
4 \sqrt{a}-a=8 \sqrt[4]{a}-4 \sqrt{a}
$$

Letting $x=\sqrt[4]{a}$ gives

$$
x^{4}-8 x^{2}+8 x=0,
$$

and since $a \neq 0 \Longrightarrow \sqrt[4]{a}=x \neq 0$, we have

$$
x^{3}-8 x+8=0 \text {. }
$$

We can observe that 2 is a solution to this equation, so we can finish by determining the solutions to $\frac{x^{3}-8 x+8}{x-2}=x^{2}+2 x-4=0$. The only positive solution is $\sqrt{5}-1$, but if $\sqrt[4]{a}=\sqrt{5}-1$, then $a$ would not be an integer. Hence, the only possible values for $a$ is 16 , in which case all three terms of the sequence are 16 .