Doppler redshift in special relativity

I came across this exercise in Elementary General Relativity by Alan MacDonald:

A source of light pulses moves with speed v directly away from an observer at rest in an inertial frame. Let $\Delta t_e$ be the time between the emission of pulses, and $\Delta t_o$ be the time between their reception at the observer. Show that $\Delta t_o = \Delta t_e + v\Delta t_e$.

Based on my understanding of special relativity, the space-time interval between two events as measured from two inertial frames of reference should be the same. Therefore,

$$\Delta t_e^2 = \Delta t_o^2 - \Delta x^2$$ $$\implies \Delta t_e^2 = \Delta t_o^2 - v^2\Delta t_o^2$$ $$\implies \Delta t_o = (1 - v^2)^{-1/2}\Delta t_e$$

which is not the same relation. What is wrong with my reasoning?

Answer

Your answer is right assuming $\Delta t_e$ is the interval between emission as measured by the emitting source itself. The given answer is right assuming $\Delta t_e$ is the time between emission as measured by the observer. It seems as though this problem is aiming at a lower level than your current understanding of relativity; you put too much thought into it.