"The answer lies in their one part in a thousand extra mass. Adding neutrons costs a little in mass, and in Einstein $E=mc^2$ implies that mass equates to energy. So it costs energy to add neutrons(due to their mass) and it costs energy to add protons(due to their electrical repulsion). This qualitatively explains why the elements have N larger than P, but not by too much.
This extra energy locked into the neutron's mass also leads to instability. As nature seeks the state of lowest energy, like water running downhill to sea level, so a neutron left to itself will eventually experience beta decay: $n\to p+e+\nu$, which converts it to the marginally lighter proton, the excess energy being transformed into the electron and neutrino. So while a neutron in a nucleus can be stabilised, if you gather too many together, they will undergo beta decay, increasing the number of protons at the expense of neutrons. Conversely trying to put too many protons together and their net electrostatic repulsion destabilises them; in this case it is possible to lower the net energy by "inverse beta decay" where one of the protons converts to a neutron:
$$ \rm p\text{ (in nucleus)} \to n \text{ (in nucleus)} + \text{positron} + \nu. $$
The net effect is when collections of N and P get too big a mismatch, beta decay or inverse beta decay moves the whole back toward the "valley of stability" where the number of neutrons N tends to exceed the number of protons P."
(The above is from a book called "Cosmic onion. Quarks and the nature of universe")
From my understanding (not too sure as I am new on this), it took much more energy to gather protons together as compared to gathering neutrons together. What holds them together is the strong force and since the strong force holding them together is the same, there will be "a lot of strong force" left over when holding just neutrons together. As nature seeks to minimise the energy state of the nucleons, some of the neutrons will decay to proton such that there will not be "any strong force left over", hence ensuring that the energy state of the necleons are minimise.
If what I say above is true, the neutron and proton "produce" equal amount of force carrier particles as is it because they themselves are made of quarks held together by the same colour charge and thus each of them produce the same amount of residual strong force?
Best Answer
The strong force has an approximate symmetry called isospin, which is a rotation in an abstract space spanned by the neutron and proton. This symmetry is broken by the Coulomb force, which only acts on protons, and by the slight mass difference between up quarks and down quarks, which makes the neutron slightly heavier than the proton.
Consider possible bound states of neutrons and protons. These are fermions, so the wave function has to be anti-symmetric. Neutrons and protons are heavy, so we can analyze the situation using non-relativistic quantum mechanics. If a bound state exists, the ground state is in an s-wave (symmetric) because the centrifugal barrier is purely repulsive. This means that the spin-isospin wave function has to be anti-symmetric. This leaves two possibilities: $$ (I=0,S=1) \;\;or\;\; (I=1,S=0) $$ where $I$ and $S$ are the total isopsin and spin of the state. Note that $I,S=0$ is anti-symmetric in isospin/spin, and $I,S=1$ is symmetric. The $I=0$ state corresponds to $$ |I=0\rangle = \frac{1}{\sqrt{2}}\left(|np\rangle -|pn\rangle\right) $$ and $I=1$ has three isospin components $$ |I=1,I_3=+1,0,-1\rangle = \left\{|pp\rangle ,\frac{1}{\sqrt{2}}\left(|np\rangle +|pn\rangle\right),|nn\rangle\right\} $$ The potential in the $I=0,1$ channels is determined by complicated interactions between quarks and gluons.
Empirically (or from numerical calculations in lattice QCD) we know that both the $I=0$ potential and the $I=1$ potential are attractive, but the $I=0$ interaction is slightly more attractive.
Note that in non-relativistic QM (in three dimensions), an attractive interaction is not sufficient to form a bound state. The potential has to have a minimum strength. Nuclear physics is fine-tuned, the empirical NN potentials are close to the threshold where bound states appear.
There is an $I=0$ bound state, called the deuteron (binding energy 2.2 MeV), but the di-neutron (and its isospin partners) fall just short of being bound. Due to the Coulomb force, there is a little extra repulsion in $pp$, but the $nn$ channel is almost bound.
This can be quantified using the scattering length. The $nn$ scattering length is about 20 fm, and the corresponding energy scale is $$ B=\frac{1}{2ma^2}\simeq 50 \, keV $$ so the di-neutron comes within 50 keV of being bound. This also means that if one could change the quark masses slightly, there would presumably be a bound di-neutron (this can be checked in lattice QCD).
It is possible in principle to have a three neutron bound state even if there is no two neutron bound state (states like that are called Borromean), but this is disfavored by the Pauli principle (not all three neutrons can be in an s-wave). The same applies to 4n bound states (there is a recent claim of a 4n resonance http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.052501).
In practice $^3H=pnn$ and $^3He=ppn$ exist, but not $3n$.
Note that neutron stars, clusters of $10^{57}$ neutrons, do exist. Of course, these objects are gravitationally bound.