Unique 1

Let positive integer $N = \sqrt{\overline{2a9b1}}$ . Then the range of $N$ is given by: $\left \lceil \sqrt{20901} \right \rceil \le N \le \left \lfloor \sqrt{29991} \right \rfloor$ or $145 \le N \le 173$ . Then $N$ is a 3-digit integer of the form $N = \overline{1mn}$ . Since the perfect square $\overline{2a9b1}$ ends with 1, $n$ can only be $1$ or $9$ and $m=4,5,6,7$ .

For $n=1$ , then $(101+10m)^2 = 10201 + 2020m + 100m^2 = \overline{2a9b1}$ . Equating the hundredth digit on both sides, we have:

$\begin{aligned} \left \lfloor \frac {20m}{100} \right \rfloor + \left \lfloor \frac {100m^2}{100} \right \rfloor \bmod 10 + 2 & = 9 \\ \left \lfloor \frac m5 \right \rfloor + m^2 \bmod 10 & = 7 \\ \implies m & = 6 \end{aligned}$

Then $N = 161$ and $N^2 = 161^2 = 25921$ , acceptable.

For $n=9$ , then $(109+10m)^2 = 11881 + 2180m + 100m^2 = \overline{2a9b1}$ . Equating the hundredth digit on both sides,

$\begin{aligned} \left(\left \lfloor \frac {18m+88}{10} \right \rfloor + m^2 \right) \bmod 10 & = 9 \\ \left(\left \lfloor \frac {9m+44}5 \right \rfloor + m^2 \right) \bmod 10 & = 9 \end{aligned}$

where $m = 4,5,6$ . For $m=4, 5, 6$ , the $\text{LHS} = 2, 2, 5 \text{ (mod 10)}$ respectively. So there is no solution.

Therefore, $N^2 = 25921$ , $\implies a^b + b^a = 5^2 + 2^5 = 25+32 = \boxed{57}$ .

minimum and maximum of $\overline{2a9b1}=y^2$ are $20901$ and $29991$ , respectively. Using a calculator, the square roots of the number are approximately $144$ and $173$ . So $y$ is either the form $\overline{1x1}$ or $\overline{1x9}$ , because the last digit go the square number is $1$ . Therefore, we need to search $2 \times 3$ (because the second digit of $\overline{1x1}$ or $\overline{1x9}$ is bounded by $4$ and $7$ ) numbers to see which has the form $2a9b1$ , when taken to the power of two. turns out the number is $25921=161$ . Finally calculate $a^b+b^a=5^2+2^5=57$