A number theory problem by Vishal S

From the equation $a^2 + d^2 = 2ad$ ,

$a^2 -2ad + d^2 = 0$

$(a - d)^2 = 0$

$a = d$ .. Equation 4

Substitute Equation 4 to the second equation $ac = d^{3 \over 4}$ ,

$dc = d^{3 \over 4}$

$c = d^{-{1 \over 4}}$ .. Equation 5

Substitute both Equation 4 and Equation 5 to the first equation $a^{b} = c^{1 \over 2}$ ,

$d^{b} = (d^{-{1 \over 4}})^{1 \over 2}$

$d^{b} = d^{-{1 \over 8}}$

Hence $\boxed {d = -{1 \over 8}}$ .

Sum of numerator and denominator is $-1 + 8$ (I'm not quite sure if one should also consider that $-7$ can also be an answer because one can put the minus sign in the bottom (leaving the denominator instead of the numerator to have the negative value) and the sum of the numerator and the denominator would have been $1 + (-8)$ .) which is $\boxed {7}$ .