What makes it occur? How do the protons and nucleus know that they have to lose mass to produce energy…?
And is the mass of a compressed Spring more than an uncompressed one??
does a body which has a greater energy has more mass than the one which has a less energy?
[Physics] Mass Defect…cause and origin
binding-energymass-energynuclear-physics
Related Solutions
To understand binding energy and mass defects in nuclei, it helps to understand where the mass of the proton comes from.
The news about the recent Higgs discovery emphasizes that the Higgs mechanism gives mass to elementary particles. This is true for electrons and for quarks which are elementary particles (as far as we now know), but it is not true for protons or neutrons or for nuclei. For example, a proton has a mass of approximately $938 \frac{\mathrm{MeV}}{c^2}$, of which the rest mass of its three valence quarks only contributes about $11\frac{\mathrm{MeV}}{c^2}$; much of the remainder can be attributed to the gluons' quantum chromodynamics binding energy. (The gluons themselves have zero rest mass.) So most of the "energy" from the rest mass energy of the universe is actually binding energy of the quarks inside nucleons.
When nucleons bind together to create nuclei it is the "leakage" of this quark/gluon binding energy between the nucleons that determines the overall binding energy of the nucleus. As you state, the electrical repulsion between the protons will tend to decrease this binding energy.
So, I don't think that it is possible to come up with a simple geometrical model to explain the binding energy of nuclei the way you are attempting with your $\left(1\right)$ through $\left(15\right)$ rules. For example, your rules do not account for the varying ratios of neutrons to protons in atomic nuclei. It is possible to have the same total number of nucleons as $\sideset{^{56}}{}{\text{Fe}}$ and the binding energies will be quite different the further you move away from $\sideset{^{56}}{}{\text{Fe}}$ and the more unstable the isotope will be.
To really understand the binding energy of nuclei it would be necessary to fully solve the many body quantum mechanical nucleus problem. This cannot be done exactly but it can be approached through many approximate and numerical calculations. In the 1930's, Bohr did come up with the Liquid Drop model that can give approximations to the binding energy of nuclei, but it does fail to account for the binding energies at the magic numbers where quantum mechanical filled shells make a significant difference. However, the simple model you are talking about will be incapable of making meaningful predictions.
EDIT: The original poster clarified that the sign of the binding energy seems to be confusing. Hopefully this picture will help:
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This graph shows how the potential energy of the neutron and proton that makes up a deuterium nucleus varies as the distance between the neutron and proton changes. The zero value on the vertical axis represents the potential energy when the neutron and proton are far from each other. So when the neutron and proton are bound in a deuteron, the average potential energy will be negative which is why the binding energy per nucleon is a negative number - that is we can get fusion energy by taking the separate neutron and proton and combining them into a deuteron. Note that the binding energy per nucleon of deuterium is $-1.1 \, \mathrm{MeV}$ and how that fits comfortably in the dip of this potential energy curve.
The statement that $\sideset{^{56}}{}{\text{Fe}}$ has the highest binding energy per nucleon means that lighter nuclei fusing towards $\text{Fe}$ will generate energy and heavier elements fissioning towards $\text{Fe}$ will generate energy because the $\text{Fe}$ ground state has the most negative binding energy per nucleon. Hope that makes it clear(er).
By the way, this image is from a very helpful article which should also be helpful for understanding this issue.
A nucleon in nuclear context is simply not the same as one in a free context. Not in mass nor in form factor. These corrections are not known in complete detail but there are parameterizations of them that are used in nuclear and particle physics experiments. In my disertation project we used a parameterization due to de Forest, which is a popular but now somewhat dated model.
Best Answer
Notwithstanding the comments that this is a silly philosophy question, I think it is a good question for precisely that reason: you know that protons cannot "know" things and therefore we must find an explanation not involving "knowledge". It sounds as though you're thinking something along the lines of everything's getting to the end of the reaction and saying "hey, we need to have less mass now", which of course you know is preposterous. So maybe it helps to think of fusion / fission as a process: he protons are in constant contact with the process, and it is their loss of energy that "drives" the process (I use the quotes because the energy loss is necessary, but not sufficient, to make the process happen). At the risk of being too colloquial, you can almost say the protons use a small piece of themselves up in completing the reaction.
(Rest) Mass is simply the property that something has if it has a nonzero energy content as measured from a frame that the particle is at rest relative to. As in the other answers, this notion is neatly expressed in the equation (for the "four-momentum's Minkowski norm"(see footnote)) $E^2 - p^2\, c^2 = m^2\,c^4$: if you are in a rest frame relative to a particle, then its momentum $p$ is nought and it has an energy content if and only if $m\neq0$. So yes, mass really does measure an energy content. So after having taken part in the process, the protons have a smaller energy content than before as measured from a relatively stationary frame, so that they have less rest mass.
But not only does the property rest mass measure energy. It also gives rise to the property of inertia, or resistance of state of motion change to external force, i.e. to the proportionality constant $m$ in Newton's second law. For instance, if we confine a quantity of light inside a perfect, lossless resonant cavity, we can show that the system's inertia increases by an amount $E/c^2$ when we confine the light, where $E$ is the light's energy. I talk about the thought experiment that shows this in my answer here. Indeed, in the same way, most of the mass in your body is owing to the massless, but confined, gluons in the nucleusses of your body's atoms.
For accuracy, I should say that particle physics thinks of many conversions as noncontinuous events: simply branches in a Feynman diagram and does not try to penetrate the "process" or think of it as a continuous evolution as I have implied. But the key idea is that everything is connected and interacting, with the transfer of energy. For example, the fusion reaction of four protons to yield helium in the Sun is thought of as three discrete events:
The proposal of this process as the source of energy in stars led to the award of the 1967 Nobel prize to Hans Bethe
Footnote: I appreciate this phrase is likely to be jargon to you at this stage - I'm not trying to be a bothersome git- I use the phrase because you might like to use it as a phrase to google on as your understanding builds and you want to know where it comes from. See here.