find numbers of n

• Let $\sqrt{n-64} = y \implies n-64 = y^2 \implies n = y^2 +64$
• Let $\sqrt{n} = x \implies n = x^2$

Now we have, $\begin{aligned} x^2 - y^2 &= 64 = 2^6 \\ (x+y)(x-y) &= 2^6 \\ (x+y)(x-y) &= \left\{\begin{array}{c}&2^6 \cdot 1 \\ &2^5 \cdot 2 \\ &2^4 \cdot 2^2 \\ &2^3 \cdot 2^3 \end{array} \right. \text{ as x+y is integer and } x+y \geq x-y \\ \\ \implies (x,y) &= \left( \cfrac{(x+y)+(x-y)}{2}, \cfrac{(x+y)-(x-y)}{2} \right) \\ &= \left\{\begin{array}{c}&\cancel{(32.5, 31.5)} \hspace{5mm} \\ &(17, 15) \\ &(10,6) \\&(8, 0) \end{array} \right. \\ \implies n = x^2 &= \left\{\begin{array}{c}&17^2 = 289 \\ & 10^2 = 100 \\ &8^2 = 64 \end{array} \right. \\ \\ \end{aligned}$ $\therefore$ The answer is $\boxed{3}$ .

There are three instances:

$1.\sqrt{64}+\sqrt{64-64}$

$2. \sqrt{100}+\sqrt{100-64}$

$3. \sqrt{289}+\sqrt{289-64}$

We just need to find the Pythagorean triples containing an $8$ and those will be the answers.