Light bulb in a series with a resistor

A light bulb is wired in series with a 132 ohm resistor and they are connected across a 120V source. The power delivered to the light bulb is 22.8 W. What are the two possible resistances of the light bulb?

I can't figure this problem out! I tried setting V=IR and P=IsquaredR equal to each other but it's not right.

The voltage the light bulb experiences is
V = (120V)*R / (132 + R)
The power the light bulb delivers is
P = V^2/R
Plug in the voltage the light bulb experiences
P = (14400V)*R / (132 + R)^2
Doing some not-so fun algebra we find that the two resistances are
R = 55.9 ohms or 311.7 ohms

I'm afraid one more piece of information is needed here, like what is the bulb RATED at? Let's assume for a moment that it's a 25 Watt bulb, (the next higher standard rating), so

Assuming the lamp's normal rating
P = 25 Watts
E = 120 Volts
I = .02083 Amperes
R = 576.1 Ohms at rated current.

P = 22.8 Watts Actual - Checks
E = 114.6 Volts across lamp
I = .1989 Amps through the circuit.
R = 576.1 Ohms (assumed)
{Twinkle twinkle little star, Power equals I-squared-R)
Restated
__ P __
I2 * R

Since checking by multiplying out gives a power of 22.795 which rounds to 22.8 this answer is correct.

Ok, let's see what happens if the bulb is rated at 40 Watts?
P = 40
E = 120
I = .3333
R = 360 Ohms

132 + 360 = 492
P = 59.256
E = 120
I = 0.4938
R = 492
Nope, it is not a 40 Watt bulb! Let's try 60 Watts?
P = 60
E = 120 Volts
I = 0.5 Amps
R = 240 Ohms

240 + 132 = 372
P = 38.709 total Watts 60 Watt bulb
E = 120
I = .3225
R = 372

P = 22.8 ( Check: 23.793 - Close but no cigar.)
E = 92.095
I = .2476
R = 372

Let's try something here...
P = 22.8
E = 92
I = 0.247
R = 372.469

So I came up with 372.47 and 576.1 Ohms. This was a strange one!