Question

F $1 \le a \le 10$ and $1 \le b \le 36$, for how many ordered pairs of integers $(a, b)$ is $\sqrt{a + \sqrt{b}}$ an integer?

Answer 1

You have such entry data: 
$1\ \textless \ a\ \textless \ 10 \\ 1\ \textless \ b\ \textless \ 36$

Consider expression $\sqrt{a+ \sqrt{b} }$. If this expression becomes an integer, then b=4,9,16,25, because then $\sqrt{b} =$ 2,3,4,5, respectively. In other cases $\sqrt{b}$ is not integer and thus the expression $\sqrt{a+ \sqrt{b} }$ also is not integer.

1. b=4, then $\sqrt{a+ \sqrt{b} }=\sqrt{a+2}$. Here a=2,7 (in other cases $\sqrt{a+2}$ is not integer). When a=2, $\sqrt{a+2}=\sqrt{2+2}=2$ and when a=7, $\sqrt{a+2}=\sqrt{7+2}=3$.

2. b=9, then a=6 and $\sqrt{a+ \sqrt{b} }= \sqrt{a+3}=\sqrt{6+3}=3$.

3. b=16, then a=5 and $\sqrt{a+ \sqrt{b} }= \sqrt{a+4}=\sqrt{5+4}=3$.

4. b=25, then a=4 and $\sqrt{a+ \sqrt{b} }=\sqrt{a+5}=\sqrt{4+5}=3$.

Answer: (2,4), (7,4), (6,9), (5,16), (4,25).