Berkeley Math Circle

Solution:
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
$$x^{3} - 8 x + 8 = 0.$$
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 value for $a$ is $16$, in which case all three terms of the sequence are $16$.