What does proving the Collatz Conjecture entail? From the get go: i'm not trying to prove the Collatz Conjecture where hundreds of smarter people have failed. I'm just curious. 
I'm wondering where one would have to start in proving the Collatz Conjecture. That is, based on the nature of the problem, what's the starting point for attempting to prove it? I know that it can be represented in many forms as an equation(that you'd have to recurse over):
$$\begin{align*}
f(x) &= 
  \left\{
    \begin{array}{ll}
      n/2 &\text{if }n \bmod2=0 \\
      3n+1 &\text{if }n \bmod2=1
    \end{array}
  \right.\\
\strut\\
a_i&=
\left\{
\begin{array}{ll}
n &\text{if }n =0\\
f(a_i-1)&\text{if }n>0
\end{array}
\right.\\
\strut\\
a_i&=\frac{1}{2}a_{i-1} - \frac{1}{4}(5a_{i-1} + 2)((-1)^{a_i-1} - 1)
\end{align*}$$
Can you just take the equation and go from there?
Other ways I thought of would be attempting to prove for only odd or even numbers, or trying to find an equation that matches the graph of a number vs. its "Collatz length"
I'm sure there's other ways; but I'm just trying to understand what, essentially, proving this conjecture would entail and where it would begin. 
 A: Adding to @Adam B.'s answer: examinig the conditions of possible cycles leads to the relations of powers of 3 and powers of 2, focusing on problems which are still not solved either.      
One can look at it in terms of approximation : what is the smallest difference between perfect powers of 3 and perfect powers of 2, relative to the magnitude of one of them or of the rational approximation of log(3)/log(2) where we find an unsolved detail in the Waring-problem (see mathworld, "power fractional parts"). Some first steps to the proof of nonexistence of cycles (in the positive integers, in the negative integers we have at least 3 additional cycles) were provided by Ray Steiner 1996 and later by John Simons and Benne de Weger who proved the nonexistence of a certain type of cycles using that rational approximation approach.     
Or one can look at the problem of cycles in terms of modular conditions, and arrive at other unsolved properties of the relation of powers of 3 to powers of 2. There is, for instance, the formulation in terms of "z-numbers" done by Kurt Mahler.   
Unfortunately, even if that 3/2-problems were solved, that would not mean that the Collatz-problem was also solved and vice versa; for instance the solution of the Waring-problem-detail would only solve the "1-cycle" problem but not the general "m-cycle" problem with m bigger than roughly 70 (using the notation of Simons/De Weger): the mentioned conditions are not including each other. (The Steiner/Simons/De Weger articles are linked in the wikipedia, a more basic, amateurish article of mine adressing these aspects a bit less cryptic can be found here )
A: The way I interpret the problem is
For there not being cycles:
That Collatz (What else? Important question) iterations change the prime factors of some arbitrary number such that future (Collatz) iterations of that number can never return to the original, or any preceding, set of prime factors. In essence, that prime factors, through at least these specific functions, get "mangled beyond return." Or that the only path to a previous set of prime factors is by inverting your previous operations, though this is probably too strong.
For no diverging numbers:
The aforementioned set of prime factors converges to a set of solely {2^n}, and thus to the empty set.
A: Proving the conjecture is equivalent to finding a measure of "simplicity" of positive integers such that repeatedly applying the function in the $3x+1$ problem eventually simplifies any input.  
That is precisely what would be entailed in any proof, but apparently it is extremely hard to find such a thing. The $3x+1$ iteration can increase all the commonly used height measures (such as ones based on the size of an integer or its prime factorization) quite far before bringing them back down, and bounding those excursions is the same problem of finding a quantity that controls the process.
By the same logic, a disproof by showing an infinite orbit exists would necessarily entail finding a complexity measure such that the iteration, on some inputs, makes the integer increasingly complex.  This is far beyond current ability for the same reason, and also because the conjecture appears to be true.
The other possibility is to find a finite cycle. Finding one does not necessarily entail more than a giant computation, but there could, hypothetically, be other methods than searching directly for a cycle.
A: Proving this conjecture indirectly would entail two things:


*

*Proving that there is no number n which increases indefinitely

*Proving there is no number n which loops indefinitely (besides the 4, 2, 1) loop
If one does these things then you have an answer to the collatz conjecture (and if you find a case of either of these things you have disproven the collatz conjecture obviously)
Of course this is just one approach that comes to mind, there are other possible methods which are beyond my own knowledge
A: The problem can be just those two points: 
1) is there a loop?
2) is there a sequence that increases without bound? 
However, another way to solve it would be to show there cannot be two distinct < families >. Up to quite high values of n, we know empirically that starting with any given n less than that value, repeatedly choosing n/2 for even n and 3n+1 for odd n gives a sequence ending in 1. 
Call this set of sequences ending in 1 the < terminate-in-1 family >. 
The task then amounts to testing if there can be a < deviant family > where either a loop or a sequence increasing without bound would amount to a deviant family. Then the challenge (in order to prove Collatz/Ulam/Thwaites correct) is to show that any other family of sequences must somewhere produce a number that is within the terminate-in-1 family. 
The terminate-in-1 family contains, and if it existed any deviant family would contain, infinitely many natural numbers each. 
A: One approach could be to use the inverse function. Start with unity and apply the inverse function $g(n)$ recursively. Visualize each positive integer as a node and each iteration as a path. To prove the conjecture, one has to show that every node will be reached eventually. To disprove the conjecture, one has to show that there exist at least one node which will never be reached. 
\begin{align*}
g(n) &= 
  \left\{
    \begin{array}{ll}
      2n &\text{if }n \bmod2=0 \\
      (n-1)/3 &\text{if }n \bmod2=0 \hspace{5pt} \& \hspace{5pt} n \bmod3=1  \\
      2n &\text{if }n \bmod2=1
    \end{array}
  \right.\\
\strut\\
\end{align*}
Note that while $f(n)$ is a one to one function, $g(n)$ is not.
See this question.
