Find the $m$ such $a_{n+1}=a^5_{n}+487$

Question

Let $\{a_{n}\}$ be a sequence of positive integers, and suppose $a_{0}=m$. Further, $\{a_{n}\}$ satisfies $$a_{n+1}=a^5_{n}+487.$$ Find $m$ so that this sequence consists of square numbers for as long as possible.

Answer 1

You have the recurrence $$a_{n+1} = a_n^5 + 487$$ And we want that atleast $a_0$ and $a_1$ are perfect squares. Let $a_1=n^2$ and $a_0 = p^2$. Then, $$a_1 = a_0^{5} + 487$$ and $$n^2 = p^{10} + 487$$ Putting $p^5:=t$, $$n^2 = t^2 +487$$ $$\implies n^2-t^2 = 487$$ Note that $487$ is prime. So, $$\implies (n-t)(n+t) = 1\times 487$$ $$\implies n-t=1, \ \ n+t = 487$$ $$\implies n = 244, \ \ t = 243 = 3^5 \implies p = 3$$ Thus, we have $a_1 =n^2= 244^2$ and $a_0 =p^2= 3^2=9$. On computing $a_2$, we find that it is not a perfect square. ($a_2=747994256939818786226663$)