Let $\frac{a_k}{b_k}$ be the $k^{th}$ convergent of the simple continued fraction of the prime $p$. Is it possible that $a_k = s^2$ and $b_k = t^2$ for some integers $s$ and $t$?
I have looked at $k^{th}$ convergents of $\sqrt{3}$ and noticed that this is not possible. Hence, I have come up with this observation. I wonder if it would lead to a contradiction.
Edit:
It turns out this is possible, as mentioned in the comments. So I would like to know for which primes $p$, $k^{th}$ convergent of $\sqrt p$ can never be of the form $\frac{s^2}{t^2}$?