No, this is exactly the wrong point here. I do not know of any theory, which would predict such strange behaviour, especially not wave theory. You just neglect constructive interference. Consider the case of an ideal detector with 100% quantum efficiency, which means that any photon will be detected. While your understanding of destructive interference would cause photons to go through undetected. Howwever at the places of constructive interference nothing would happen. Any photon would be detected anyway due to the 100% QE, so there are just two conclusions left:Not true. This is what QM predicts, not wave theory. In wave theory, there are just as many photons hitting the dark fringes as there is hitting the bright fringes, only they're not setting off the detector because in these exact positions, they are interering destructively. Therefore, the detection count is significanlty lower compared to firing rate in wave theory.
a) The constructive interference will lead to an increase in photon detections, which cancels the decrease of the destructive interference area. Therefore no decrease in the total number of photon counts will occur.
b) There are no effects of constructive interference at all. The number of photons detected at the areas of wannabe constructive interference equals the number of photons detected with no double slit present. This is either a violation of conservation of energy or (if you think that the wave will just travel through the screen and be detected somewhere else - however this might happen) easily disproved by putting a second detector screen after the first one, which should then show bright spots, where there have been areas of destructive interference at the first screen. You can easily build your own double slit at home and show, that the second option is plain wrong.