I am aware of the fact that the squares don't contain an infinite arithmetic subsequence, but I was wondering if the squares contain an infinite geometric sequence.
In other words, does there exist an infinite geometric sequence whose terms are all squares?
On
Let $a_1 = m^2$ and $a_2 = n^2$ while $a_2 = r*a_1$ so $r = (\frac nm)^2$ a square of a rational. If $r$ in lowest terms is $\frac jk$ then for $a_{n+1} = (\frac jk)^n*a_1$ to be an integer $k^n$ must divide $a_1$. But that is true for all values of $n$ and the only way that is possible is if $k = 1$.
So if $a_1 = m^2$ and $r = j^2$ then... well that'll do it. By induction if $a_n = l^2$ for some integer $l$ then $a_{n+1} = r*l^2 = j^2 *l^2 = (jl)^2$ is a prefect square.
Ex: $a_1 = 9=3^2$ and $r = 4$ then $a_{n+1} = 4^n*9= (2^n*3)^2$ for all $n \ge 1$.
Set $a=1$ and $r=4$, so the geometric progression becomes $$1,4,16,64,256,\dots$$ which is equivalent to $$1^2,2^2,4^2,8^2,16^2,\dots$$