Why don't neutrons in a nucleus decay? When I asked here why neutrons in nucleus (with protons) don't decay I was told that it would require energy for the neutron to decay, it wouldn't give energy. And since that wasn't really what I wanted to hear, since I already knew that, I'm now asking a similar question.
I know that strong force holds protons and neutrons together in nucleus, but how does that effect a neutron in a way that it doesn't decay? 
 A: 
Why are there neutrons in the nucleus of an atom?

Protons are positively charged. The electromagnetic force between protons is repulsive even though the residual strong force from the quark content is attractive, the combined potential is not attractive so as to produce a bound state.  
Let us take the simple system of two protons with their repulsive potential. The strong nuclear potential added to the electromagnetic repulsion does not give bound energy levels that could accommodate the quantum numbers of the two protons. The addition of the charge-neutral  neutron, which,( even though a composite of charged quarks,) has an overall  strong force attraction overcomes the repulsive potential and deepens the the total  potential so that it can make a potential well to trap in a bound state the three nucleons, He3, an isotope of the alpha nucleus, in appropriate quantum number states.  The Pauli exclusion principle  also plays a role . See  here.

I know that strong force holds protons and neutrons together in nucleus, but how does that effect a neutron in a way that it doesn't decay? 

If it were not for the weak interaction the neutron would also be a stable particle, as is the proton. Thus the answer to "why" has to start with the basic decay of the neutron which depends on the quark interactions with the weak force.
The proton is composed out of (u u d) quarks and the neutron out of (u d d). In both the effective mass measured for quarks and in the  particle table the up quark has the smallest mass. ( these masses depend on models, but the standard model is very successful in describing elementary particle interactions)
The up has a mass of 2.3 MeV, and the down 4.8. The up is stable as it cannot decay into a lower mass quark. To change an upquark to a down quark , at least 2.5 MeV of energy has to be provided.

The neutron has two down quarks, and they are free to decay to an up quark because it is energetically favorable, and thus it decays to aproton  with a lifetime of 900 seconds when free. A proton's single down could  also decay , but there is no stable (u u u) lower energy than the proton state within the standard model interactions. Proton decays need new mediating particles and are predicted in various extended models, and have not been detected experimentally up to now.
A nucleus is a many body problem and its quantum mechanical solutions are approximated by the shell model for example, where an effective potential keeps the nucleons, protons and neutrons, in a bound energy level. Thus in a stable nucleus the neutron will be sitting at an energy level 2.5 MeV below the escape threshold  and thus will not decay. In an unstable nucleus the effective potential extends above the zero line and there exist energy levels for the neutron to occupy but also a probability since it is energetically possible, to decay. In a simple quantum mechanical model it could tunnel out and decay, as long as the energy level it is bound in is above 0.
See this link for a shell model description on how the proton and neutron see the ( approximate) potential well, strong for neutron, strong and Coulomb for proton.

One should keep in mind that by pion exchange within the nucleus a proton can turn into a neutron and vice verso, in the details of the interactions. Thus a proton sitting in an energy level above the zero energy has a probability to turn into a neutron by pion exchange which can then decay.
As a nucleus is a many body state, quantum numbers also play a large role, and the Pauli exclusion principle. For example the decay will be inhibited if the resulting proton from the neutron decay has no allowed energy level to occupy.
So the answer to " but how does that effect a neutron in a way that it doesn't decay?" is that it is all dependent on the effective many body potential . Neutrons in energy levels below escape threshold, bound, will not decay, in nuclei where there are energy levels above threshold they have a probability of decay since it is energetically possible, if the quantum number conservations allow it..
A: There is actually a pion exchange between nucleons.  This is the basis for the residual strong force.
When a proton and neutron interact, one of the interactions we get is
$$\rm proton^+ +neutron \rightarrow proton^++pion^-+proton^+ \rightarrow neutron+proton^+$$
So the proton becomes a neutron and then the neutron becomes a proton by transferring a negative pion.

Edits:
@Rob Jeffries:  That's fair, my answer didn't really answer.

The proton and neutron aren't identical particles and so aren't limited by Pauli Exclusion between each other.  They can have the same spin state and for some reason having parallel spins gives the combination a lower energy and so more stability.  
When I look for why parallel spins have a lower energy, I just see it stated as a fact or in better references, an equation that gives the details of the energy difference is given. 
What I haven't seen is a physical explanation for why parallel nuclear spins have lower energy.  
Here's my hunch (which I will try to verify).  If it's correct then it's probably out there somewhere already.  If it's wrong, it's because I just thought the idea through myself.
When a proton and neutron have all the same quantum numbers (and so their spins are parallel), then when a neutron transfers a negative pion to the proton, the final state is identical to the initial state.  This means there are two different paths from the initial state to the final state: doing nothing and passing a negative pion both give you the same final state.  
In quantum when two different events are possible because they both give the same initial and final state there is an interference term.  For instance, in the collision of two identical atoms with identical spins states, colliding and not colliding can give the same final state, and so the two possibilities interfere.
In the case of a proton and a neutron with antiparallel spins, transferring a negative pion would give a different final state than the initial state.  It could transfer back to then give the same final state, but (roughly speaking) every step a process needs makes the contribution of that process less important.
In the case of a diproton or dineutron, the transferring a neutral pion would also give the same final and initial state.
So I further suspect (again, all on me if it's wrong) that while the proton-neutron parallel pair has dominant exchanges of both neutral pions negative pions, the diproton and dineutron only has neutral pions giving the proton-neutron pair about twice the binding. 
So the two key ideas I'll try to track down are:
(1) Does a process having identical initial and final state make it more important?
(2) Does having twice the number of similar interactions mean that there is about twice the binding energy involved?
A: In layman's terms, if I may.   Neutron decay is possible through a rather unexpected process called the weak force or weak interaction, using the force carrying W or Z bosons, which are very energetic with a mass some 80 or 90 times the mass of the Neutron.    How that's possible is explained elsewhere and not really important, but the point is that the Neutron doesn't decay easily.   It takes 10 minutes or so, which, for a quantum state, is nearly an eternity.  
When the Proton and Neutron bind in the deuterium nucleus or deuteron, they are very close.   The diameter of a proton, about 0.81 ficometers (fm) and the diameter of the deuterium atom, about 2.1 fm.   Source.   This means that the space between the Neutron and Proton in the deuteron isn't much more than the space between the valence quarks within the Proton or Neutron.
So in that close marriage, the Neutron isn't able to destabilize, because the proton is holding it tight.    (this doesn't really say why, I admit, only that the neutron is in a very different environment when in the Nucleus vs alone), and it's only meant as a kind of layman's explanation.  I can delete if it's too far off the mark.
