Resistors or simply adding more LEDs (since they make fine resistors too).

No, LEDs do not make fine resistors. They make fine LEDs, that's all. At about 1.5V forward voltage per LED, 6 LEDs in series would run from a 9V battery without a resistor. A standard LED will run at between 1.2V and 1.8V depending on colour and the selected drive current. In your case (1 LED + 9V battery)you need to lose about 7.5V at a current of (typically) 0.02A, so the nearest series resistor would be 390 Ohms at a power rating of 0.25W. Remember not to solder the legs of the LED close to the body, unless you are very quick, or you will 'cook' the led permanently. Polarity of the led is determined by the smaller internal electrode (which is usually the positive anode) or by the flat on the flange of the LED (which is usually the negative cathode). Accidentally reversing the polarity of the LED on a supply of more than 5V could 'pop' it permanently. Always check the data sheet if in doubt. Blue LEDS need a much higher forward voltage (Typ. 4.5 to 5V) and used to be sensitive to static and reverse polarity damage. LEDs with a built-in flasher IC will usually also need a parallel zener diode to limit the LED voltage during the 'off' cycle, or they will latch up or pop. A 4.7V or 5.6V 0.25W diode usually works well.

here is a formula table for you to better understand electronic calculations & other tables

YOU SHOULD BE USING 2 1.5V BATTERYS & ABOUT A 75 OHM RESISTOR TO 100 OHM RESISTOR IN SERIES. THIS WILL LAST A LONG TIME CONTINUIOSLY ON