How do you solve for mass (without assuming it is zero) E^2 = (pc)^2 +(mc^2)^2

Well as a general rule of thumb i've come to recognize is that to get zero you either deduct the exact sum of the parts or multiply by zero. By my calculations there is no deduction that is happening equal to the sum of the parts and you are certainly not multiplying by zero, so light must have mass. Short answer is i'm trying to figure out the mass of light with a wavelength of 685

But a "photon of light" DOES NOT have any mass!! That is the reason light travel at its well-known speed, which is the "maximum" speed possible. A photon may have "wavelength" or "frequency" and , of course, "energy", but no "mass".

Then perhaps you can demonstrate the multiplication of zero or the deduction of the total sum of it's parts using the formula provided and back up this claim.

Can you explain how you switched the sides of the m, E, and most importantly how you came upon a minus sign for that nifty trick?

Honestly, could you just show the entire process, I'm still confused as to how you divided by c^2 without dividing the other side as well. Plus I just don't get where the minus comes from.

1.  Subtract (p.c)²  from both sides. to get  E² - (p.c)² = (m.c²)² 

2.  Take the square root of both sides to get \(\sqrt{E^2-(p.c)^2}=m.c^2\)

3. Divide both sides by c² to get \(\frac{\sqrt{E^2-(p.c)^2)}}{c^2}=m\)

4. Substitute for E and p using the expressions given.

Awesome thank you so much. This information is hard to find. I guess not too many people are interested in it, because it's a fact that light is always moving and hence doesn't have a rest mass.