If $1 \le a \le 10$ and $1 \le b \le 36$, for how many ordered pairs of integers $(a, b)$ is $\sqrt{a + \sqrt{b}}$ an integer?
\(1 \le a \le 10$ and $1 \le b \le 36 \)
\(\sqrt{a + \sqrt{b}} \)
b is only an integer when b = 1, 4, 9, 16, 25 , 36
The possible pairs of (a, b) are
(1, 9)
(2, 4)
(3, 36)
(4, 25)
(5, 16)
(6, 9)
(7, 4)
(8, 1)
(10, 36)
So.....nine possible ordered pairs
+612 
