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The definition of the arithmetic-geometric mean ($\text{AGM}$) for two numbers $a_0$ and $b_0$ is to define two sequences $a_n$ and $b_n$ as $a_{n+1}=\frac{a_n+b_n}{2}$ and $b_{n+1}=\sqrt{a_n b_n}$, with $\text{AGM}(a_0,b_0):=\lim_{n\to\infty} a_n=\lim_{n\to\infty} b_n$. So for example, the first few steps in computing $\text{AGM}(1,49)$ are as follows: $a_1=25$, $b_1=7$, $a_2=16$, $b_2=\sqrt{175}\approx 13.23$, etc., with the eventual result being that $\text{AGM}(1,49)\approx 14.58$. My issue is that by the second iteration we already have a non-integer term, and from the third iteration on neither term is an integer. By algebraic manipulation (for my example choice of $a_0=1$, which seems to me a reasonable starting point but certainly may not be optimal), $b_2=\sqrt[4]{\frac{b_0(1+b_0)^2}{4}}$, and it's tough for me to see when (if ever) this is an integer (avoiding the trivial case of $b_0=1$ and more generally $b_0=a_0$). But this is just one case of my more general question:

Is it possible, for some $a_0, b_0\in\mathbb{N}^+$ with (say) $a_0<b_0$, for arbitrarily many terms of both $a_n$ and $b_n$ to be integers? If so, is there a closed form for the family of least$^1$ $a_0$ and $b_0$ for which the maximal iteration at which both terms are integers is greater than for any lesser pair (for example, the least pair $(a_0, b_0)$ for which all terms up to $a_3$ and $b_3$ are integers)? And if not, what is the maximum number of iterations for which all the terms up to that number of iterations are integers and what is the least associated pair $(a_0,b_0)$?

For example, the least $(a_0,b_0)$ for which $a_0,b_0,a_1,b_1$ are all integers is $(1,9)$, giving $a_1=5$ and $b_1=3$ (and also, as it happens, an integer for $a_2$, which $=4$). My algebraic manipulation for the example of $b_2$ in the case where $a_0=1$ above suggests to me that directly working backwards from $a_n$ and/or $b_n$ to find the least possible $(a_0,b_0)$ is not a particularly fruitful avenue in general, but I could of course be missing something or just plain wrong, and apart from that I haven't found any ideas on which to attempt to follow up. Anything anyone could contribute- partial, whole, a pre-existing answer- would be greatly appreciated. Also, I can clear anything up if it's unclear or poorly stated. Thank you!

$^1$ For some definition of "least"- perhaps having the least $b_0$ or $\text{AGM}(a_0,b_0)$ would be a reasonable metric by which to judge.

Lieutenant Zipp
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What I think is that what you've found is the best we can do. In other words, we can only have all $(a_0, b_0)$ and $(a_1, b_1)$ are rational, but not upto $(a_2, b_2)$ (when $a_{0} \neq b_{0}$).

Here's my argument. By dividing terms by $a_{0}$, we can assume that $a_0 = 1, b_0 = b$ and try to find rational solution instead of integer solution (which are equivalent). To have $(a_1, b_1) = ((1+b)/2, \sqrt{b}) \in \mathbb{Q}^{2}$, we should have $b = q^{2}$ for some $q \in \mathbb{Q}$. Then we have $a_{1} = (1+q^{2})/2, b_1 = q$ and $(a_2, b_2) = ((1+q)^2 / 4, \sqrt{q(1+q^2)/2})$. For $b_{2} \in \mathbb{Q}$, we have $r\in \mathbb{Q}$ such that $$ r^{2} = \frac{q(1+q^{2})}{2} \Leftrightarrow \left(\frac{r}{q}\right)^{4} -1 = \left(\frac{1}{2}\left( q - \frac{1}{q}\right)\right)^{2} $$ so there may exists a nontrivial solution for $y^{4}-1 = x^{2}$, and this contradicts to the fact that $z^{2} = x^{4} - y^{4}$ has no nontrivial integer solution.

Seewoo Lee
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