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I must confess that I don't fully understand the wording of this problem... I was lucky enough to stumble upon the answer.

First we note that $a_{n+1}$ is a perfect square, say, $a_{n+1}=p^2$ . Then $a_{n+2}=(p-1)^2, a_{n+3}=(p-2)^2,$ etc. Thus, if $a_{33}=1^2$ then $a_2=32^2=1024$ . When the problem talks about $m$ being "minimum and greater than 1", I read that to mean that $m=2$ . Thus $B=a_2=1024$ and $A=33^2=1089$ , with $A+B=\boxed{2113}$ . I don't understand the condition about $A-B$ being a certain maximum though... maybe the author can explain.

Consider $a_{n+1} = \left( \left \lfloor \sqrt{\left \lfloor a_n \right \rfloor - \left \lfloor \sqrt{a_n} \right \rfloor} \right \rfloor \right)^2$ $\implies \sqrt{a_{n+1}} = \left \lfloor \sqrt{\left \lfloor a_n \right \rfloor - \left \lfloor \sqrt{a_n} \right \rfloor} \right \rfloor$ . Since the RHS is an integer, this means that $a_n$ are perfect squares for all $n$ . Let $b_{34-n} = \sqrt{a_n}$ . Then $b_1 = \sqrt {a_{33}} = 1$ , $\sqrt{a_2} = b_{32}$ and $b_n = \left \lfloor \sqrt{b_{n+1}^2 - b_{n+1}} \right \rfloor$ . We note that $b_{n+1} > b_n$ . Let us assume $b_{n+1} = b_n+1$ then:

$\begin{aligned} \left \lfloor \sqrt{b_{n+1}^2 - b_{n+1}} \right \rfloor & = \left \lfloor \sqrt{(b_n+1)^2 - b_n-1} \right \rfloor \\ & = \left \lfloor \sqrt{b_n^2 + b_n} \right \rfloor \\ & < \left \lfloor \sqrt{b_n^2 + b_n + \frac 14} \right \rfloor \\ & < \left \lfloor b_n + \frac 12 \right \rfloor \\ & = b_n \\ \implies b_{n+1} & = b_n + 1 \end{aligned}$

Now, if we assume $b_{n+1} = b_n+1$ then:

$\begin{aligned} \left \lfloor \sqrt{b_{n+1}^2 - b_{n+1}} \right \rfloor & = \left \lfloor \sqrt{(b_n+2)^2 - b_n-2} \right \rfloor \\ & = \left \lfloor \sqrt{b_n^2 + 3b_n+2} \right \rfloor \\ & > \left \lfloor \sqrt{b_n^2 + 2 b_n + 1} \right \rfloor \\ & > \left \lfloor b_n + 1 \right \rfloor \\ & = b_n + 1 \ne b_n \\ \implies b_{n+1} & \not > b_n + 1 \end{aligned}$

Then we have $b_{n+1} = b_n+1$ . Since $b_1 = 1$ , then $b_n = n$ , and

$\begin{aligned} \left(\left \lfloor \sqrt {a_2} \right \rfloor +1 \right)^2 + \left(\left \lfloor \sqrt {a_2} \right \rfloor \right)^2 & =\left(\left \lfloor b_{32} \right \rfloor +1 \right)^2 + \left(\left \lfloor b_{32} \right \rfloor \right)^2 \\ & =33^2+32^2 = \boxed{2113} \end{aligned}$