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

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.

• What have you tried? Does $m$ have to be square? There will be very few choices where both $a_1, a_2$ are square. – Ross Millikan Jun 15 '16 at 5:15
• Where does this question come from? I don't see a case where you get even one square out of this except for $a_0=9$, $a_1=59536=244^2$. – mjqxxxx Jun 15 '16 at 5:16
• @mjqxxxx,some years ago open problem – function sug Jun 15 '16 at 5:30
• @functionsug For a far from well-known open problem, I think you should put up rather more detail. Any references? What is the current record-holder, eg has anyone found a sequence better than the $3^2,244^2$ example? Has anyone found any other examples at all. A quick computer search up to $100000^2$ failed to find any. – almagest Jun 15 '16 at 5:48

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. ($a_2=747994256939818786226663$)
• Or you can show $a_{2 } \equiv 3$ mod 4. So it is not a square. No need to compute. – N.S.JOHN Aug 12 '16 at 17:38