One way to look at it is how much resistance is the wire adding to the circuit?
- 8 AWG = 628 µΩ per foot (2.060 mΩ per m)
- 10 AWG = 999 µΩ per foot (3.276 mΩ per m)
400 feet of 10 AWG (200 feet for supply, and another 200 for return) has a resistance of 399.6 mΩ. 600 feet of 8 AWG (300 each way) has a resistance of 376.8 mΩ. Thus the total resistance for the mixed-wire scenario is 776.4 mΩ.
1000 feet of 8 AWG (500 each way) would have a resistance of 628 mΩ.
In theory if the pump draws 12.3 A, the mixed-gauge wire will drop 9.55 V. If it were all 8 AWG, the voltage drop would be 7.72 V. The pump won't have any problems with 230-232 V.
Power dissipation (loss) in the wires is I²R. The 10 AWG wire would lose about 60.5 watts (or 151 mW per foot). The 8 AWG would lose about 57 watts (95 mW per foot). Total power lost is 117.5 W. Contrasted with entirely 8 AWG is 95 W (95 mW per foot). Switching to all 8 AWG would save you about 22.5 W of lost power.
Given the cool temperature of the ground and the long distance it is spread across, I don't think you would need to worry about exceeding any temperature ratings of insulation.
The reality however is that the pump is an inductive load, so it's not going to be able to pull its maximum current with the long run of wire. Ignoring startup or stall current, you could think of it as being a resistive load of (240/12.3) 19.5 Ω. Combined with a resistance of, let's call it 1 Ω, means the current will be a little less at (240/20.5) 11.7 A.
I am not an electrician, so check with one for any local codes you have to meet. I would say based on my basic calculations that you'll probably want to weigh the cost of ~22 watts of wasted power against the cost of larger gauge wire. How often and how long the pump runs (its duty cycle) at maximum capacity will be key.