Sorta, fresh alkaline AA are supposed to be at 1.5V(Ni-mh rechargeables are 1.2V, but that's not important right now). So two AAs would be 3 volts and that's probably enough voltage to turn this on, even though the stated minimum is 3.2, it'll just be a tiny bit dimmer. If a AA has a 2000mA capacity (they often have more than that, and don't confuse my use of the word "capacity" with capacitance, which is a completely different issue), then it can source 2000mA total before it's dead. The battery can however provide more than 2000mA of current. When we measure Amps, we measure how much electricity goes through a medium per hour. So, if we drew 2000mA from our battery, it would last 1 hour. However, if we drew 4000mA from our battery, we could only do that for 30 minutes. Likewise we could draw 1mA, and it would last 2000 hours.

When we use two batteries, we have a choice: do we hook them up in parallel, or in series. Now our AA battery will have an upper limit to how many mA we can draw from it at once, say 6000mA. If we hook them up in series(connect the + of one battery to the - of the other and use them like one battery) that limit is still 6000mA. But the voltage went up from 1.5V to 3V, and our total mA capacity went up too, from 2000mA to 4000mA. So now if we draw 2000mA from our two battery cell, it will last for 2 hours.

If we hook them up in parallel, we connect the + of both batteries to each other and the - of each to each other. Then, to use our cell, we draw from the connections between the batteries. In this setup, the voltage stays at 1.5V, and the total capacity is still at 4000mA, but our upper limit went up to 12000mA (6000 from one battery, and 6000 from the other).

Since we want 3V, we'll hook our batteries is series. The stated max current for these leds is 20 mA, this is much more important than voltage, because Leds have an interesting property that they will let any amount of current through them(that entire upper limit), even if it destroys it, so we must externally limit the current.

To limit current, we use resistance, and Ohm's law: V=IR. This is pretty simple once you get the hang of it. V stands for voltage, I for current(in Amps), and R for resistance. When we want to limit the current, we put in our voltage and target current and it gives us the resistance we need to get that current.

V=IR
3=(0.02)R
R=150

The unit for resistance is ohms, so our answer is 150 ohms. Then to limit the current to exactly 20mA, we use a 150 ohm resistor. Now, resistor values are pretty standardized, and you'll have a hard time finding a 150 ohm. To much current will kill the led, so it's better to use a higher ohm resistor than a lower one, so a 200 ohm resistor would be a good start(it'll still be very bright).

Running a single led off a battery is easy then, but running more gets slightly complicated. Leds have another interesting property called a voltage drop. If the Vdrop is 2V, all that means is that the voltage will have fallen by 2V on the gnd side of the led. On our setup, a Vdrop of 2 would leave only 1V for any other leds, which is not enough. So we have to wire them in parallel. In parallel, we wire each led with its own resistor to our 3V battery, completely independant of the other leds. In this setup, each led is getting 3V, and drawing its 20mA from the battery. So 10 leds, in parallel, with a 200 ohm resistor would draw 15mA each, 150 total, from our 3V battery cell. Since our battery has a 4000mA capacity, this would run for 26.6 hours.(of course, alkaline batteries fall off as they die, so we'd have to estimate down a bit)

Hope that explains everything! :)

They still require a resistor. If you are running them off of batteries without a resistor, then you are relying on the internal resistance of the batteries and the LED to limit current, a dangerous way to run LEDs.

So they may work on a Lithium coin cell, or two AA batteries, but burn out on a Lithium Ion rechargeable or a couple of C cells.