Well, virtual photons can have mass.

This can be verified by calculating the invariant square of a virtual photon's four vector in a suitable electrodynamic process, such as pair annihilation. During that time, since the four-momentum is conserved, the photon will appear to have mass.

of course, all this is cheating, in a sense. Such a particle is virtual by its very nature and could never be observed.
Personal pet peeve: I hate when they claim such things can be "explained" (or even just "understood") if you invoke the uncertainty principle.
Yeah, sure. But I don't feel comfortable arguing physical basic principles on account of that.

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A photon at rest?
That doesn't seem possible.
Don't go all "it's own rest frame" on me -- the principle you want to invoke (every observer is equal) is nice, but you can't use it if it conflicts with the principle of seeing light always move at c. Which means, seeing photons always move at c.
So said photon (at rest) would need to see itself moving with c. That's a contradiction.
I don't think there's such a thing as a rest frame for a massless particle (since that'd mean it moves with c).

Now, Joey is still right in saying that photons don't have mass. They also don't have mass if they move (which, uh, they always do). That does not mean photons are unable to have momentum. In fact, things get easier for them. E=p*c implies p = E/c. There you go! No need to invoke mass.
In case you want to use this for your homework, note that E=p*c is only the special case of relativistic energy-momentum in case of mass=0. E=sqrt(m^2*c^4+p^2*c^2) gives you the above if you set m=0.