Fung-sion

From the given question, for coordinates $(x,y)$

$16x^{4} - 8x^{2} + 16y^{2} - 16y +5 = 0$

Let $a = 16y^{2} - 16y + 5$ ........(1)

Hence, the equation becomes,

$16x^{4} - 8x^{2} + a = 0$ ......(2)

The roots of the above equation exists only if the discriminant of this equation is positive

Hence, $(8)^{2} - (4)(16)(a) \geq 0$

$\therefore a \leq 1$

$\therefore 16y^{2} - 16y + 5 \leq 1$

$\therefore16y^{2} - 16y + 4 \leq 0$

$\therefore 4y^{2} - 4y + 1 \leq 0$

$\therefore(2y - 1)^{2} \leq 0$

But square of any value is a positive value

Hence, $2y - 1 = 0 \implies y = \frac{1}{2}$

Putting the above value of $y$ in (1), we get,

$\therefore a = 1$

Putting $a = 1$ in (2) we get

$16x^{4} - 8x^{2} + 1 = 0$

$\therefore(2x - 1)^{2}(2x + 1)^{2} = 0 \implies x = \pm \frac{1}{2}$

Hence, there are only two order pairs $(\frac{1}{2} , \frac{1}{2})$ and $(- \frac{1}{2} , \frac{1}{2})$ which satisfies this equation

Hence the answer is $\boxed{2}$

Hint : $16x^{4} + 16y^{2} + 5 = 8x^{2} + 16y$ can be written as

$(4x^{2} - 1)^{2} + (4y - 2)^{2} = 0$