TITLE: Intuitive understanding of the uniqueness of the Fundamental Theorem of Arithmetic.
QUESTION [16 upvotes]: Basically I am trying to understand why Fundamental Theorem of Arithmetic (FTA) exists, i.e why a natural number cannot be factored primely in two or more different ways.
There are two proofs given on the wikipedia page for the uniqueness,  

Via Euclid's lemma    2. Elementary proof.

The second one, it is a proof by contradiction. The underlying reason behind this proof is a combination of the two facts,

That there must exist a smallest number which can be uniquely
factorized--kind of presumption of FTA's existence for a finite
number of natural numbers--.
A prime number cannot be factored. (The
contradiction that we get at last is $p_1(k+1)=q_1$ but $q_1$ was
assumed to be prime.)

The first fact makes sense to me as: We can show this for a finite number of natural numbers by brute-force, e.g. we can show that the numbers 1 to 15 can be uniquely factorized by making all the possible combinations of numbers less than 15.
My main problem in this elementary proof is the second fact. It doesn't give me an intuitive satisfaction which it could give if it were not the contradiction achieved but the starting point of our proof. The discoverer of this elementary proof must have it the other way round, i.e without contradiction. He should have used the fact $p_1(k+1)\neq q_1$ to prove the uniqueness of FTA(for natural numbers only). I don't know if he really had but is this possible?
The intrinsic fact (precisely the definition of a prime) that a prime cannot be factored should somehow directly imply the uniqueness of FTA. If there is some direct mathematical proof like that then this would give me a lot of satisfaction.
For a complete understanding of the first proof, I studied Euclid's lemma--which needed Bezout's identity--which further needed the concept of euclidean division. I've read all the stuff carefully but I am still not satisfied with them. I cannot grasp them for some reason, which I don't know.
How a simple Euclid division process can imply the Bezout's identity? That is what is the underlying reason (intuitive)?
In Bezout's identity we see that any number of form $ax+by$ is of the same form as the reminder obtained by dividing $a$ (or $b$) with $ax+by$ and the reminder should be 0. This is the kinda of the underlying reason. But this is not quite satisfactory to me. It's a weird fact! How the discoverer of the proof of Euclid's lemma would know that the fact that $ax+by$'s least value is $\gcd(a,b)$ implies Euclid's lemma? Is there some intuitive way to understand the chain,
 Euclid's division $\implies$ Bezout's identity $\implies$ Euclid's lemma.
 Or any intuitive way to understand directly why Euclid's lemma exists.
I don't know how to put it in words. I am not grasping all these things. All these mathematical proofs are just a way to justify the truth but they do not provide the answer to "Why is it the way it is?"
Please tell me some exhaustive intuitive explanation of some kind so that I could grasp all this stuff.
Thank you.

REPLY [4 votes]: Here are my two cents on this question. AnatolyVorobey has pointed out the ideas of the second proof with a better explanation than the one given in Wikipedia (just a remark, that proof was discovered independently by Lindemann).
First, I will focus on the usual proof of FTA. I want to remark that the chain that you wrote: "Euclid's division $\implies$ Bézout's lemma $\implies$ Euclid's lemma" is not necessary. We only need Euclid's division and the algorithm that carries his name, i.e, the euclidean algorithm. In other words, we only need the existence of GCDs for every pair of positive integers.
To prove this we need the identity: $\gcd(ac,bc)=c\gcd(a,b)$. The above result is very well-known so I'll just skip the proof. Now, going to the main result we seek: given a prime number $p$ and two positive integers $a,b$ such that $p\mid ab$, we have to prove that $p\mid a$ or $p\mid b$.
If $p\mid a$ we're done. Otherwise $p\not\mid a$, and this is the key idea here: the fact that $p\not\mid a$ implies $\gcd(p,a)=1$ and this is true because the only factors of $p$ are $1$ and $p$ itself. On the other hand, clearly $p\mid pb$ and by hypothesis $p\mid ab$, hence by the above identity and the universal definition of $\gcd$ (known also as Euclid's porism) we have $$p\mid \gcd(pb,ab)=b\gcd(p,a)=b.$$ Hence, we deduce that $p\mid b$.
Nowhere we've used Bézout's identity. So as you can see we only need the existence of GCDs and one identity to prove Euclid's lemma.
But, there is something else to say. Probably the two proofs you indicated are the most known (being the first one the most known in general), however there is another less known proof of FTA related to the uniqueness part. This third proof doesn't use Euclid's lemma, instead uses a lemma known as "Four number lemma" (also known as Euler's vierzahlensatz, but actually known since the time of Euclid). 
More exactly, this third proof uses a generalized version of the four number lemma. First, let's see what this lemma says: 

Given four positive numbers $a,b,c,d$ such that $ab=cd$, then there are four positive numbers $m,n,p,q$ such that $$a=mn,\; b=pq,\; c=mp,\; d=nq.$$

In other words, we can refine $a,b,c,d$ to another four numbers such that $a:c=d:b$. Now, the generalized version of the above lemma says this:
Given positive integers $p_1,\ldots, p_m, q_1\ldots, q_n$ such that $p_1\cdots p_m=q_1\cdots q_m$, then there are positive integers $x_{ij}$ ($1\le i\le m$, $1\le j\le n$) such that $$p_i=\prod_{j=1}^n x_{ij},\;\; q_j=\prod_{i=1}^m x_{ij}.$$
The generalized four number lemma can be proved by using double induction. I won't proved here because is somehow tedious to prove but you can check the first reference for the proof. What I'll proof is the four number lemma.
Proof of the four number lemma: We'll prove it by strong induction on $a$. If $a=1$ we have $b=cd$, so we can take the numbers $m=n=1$, $p=c$ and $q=d$. Now, let's suppose the lemma is true for every positive integers less than $a$. By Euclid's division applied to $d$ and $b$ we can find positive integers $q,r$ such that $d=bq+r$. Replacing we get $$ab=c(bq+r)=bcq+cr$$ $$\implies b(a-cq)=cr.$$
Now, since $a-cq<a$ by the hypothesis of induction we cand find positive integers $x,y,u,v$ such that $b=xy$, $a-cq=uv$, $c=xu$ and $r=yv$. Then $a=cq+uv=xuq+uv=u(xq+v)$ and let's put $xq+v=w$. Therefore we find that $a=uw$, $b=xy$, $c=ux$ and $$d=bq+r=xyq+yv=(xq+v)y=wy.\;\;\; \blacksquare$$ 
Finally let's give the promise third proof of the uniqueness part of FTA. Let's suppose that we can write a positive integer $a$ as a product of primes in two ways: $$a=p_1\cdots p_m=q_1\cdots q_n.$$ 
Then by the generalized four number lemma we can find positive integers $x_{ij}$ ($1\le i\le m$, $1\le j\le n$) such that $$p_i=\prod_{j=1}^n x_{ij},\;\; q_j=\prod_{i=1}^m x_{ij}.$$ 
Now, as the $p_i$'s are primes, there is exactly one $x_{ij}$ that's equal to $p_i$ and the other factors are $1$'s, so there are exactly $m$ factors greater than $1$. Analogously, seeing the factors of the $q_j$'s we deduce that there are $n$ factors greater than $1$. Thus $m=n$ and the terms $x_{ij}$ that are greater than $1$ give a correspondence (after rearranging if necessary) between the $p_i$'s and the $q_i$'s given by $$p_i=x_{ij}=q_i.$$ 
Hence, we have proved the uniqueness part of FTA. 
Comments: i) I think the four number lemma is a very intuitive result because it tells you that two factorizations of a positive integer have a common refinement  and this result is, as we have seen, very close to the uniqueness part of FTA.  
ii) In the first reference below there is a nice geometric proof of the four number lemma.
iii) The four number lemma can be used to prove a stronger result than Euclid's lemma, namely: if $a\mid bc$ and $\gcd(a,b)=1$, then $a\mid c$. Again, see the first reference below.
iv) It turns out there is another way to prove the uniqueness part of FTA using the four number lemma. Namely, we don't need the generalized version of the lemma above, just only the classical lemma itself. The idea is to suppose there are numbers with non-unique factorization so by the well ordering principle there is the smallest number with non-unique factorization, let's say $a$. Thus we can write $a=px=qy$ with distinct primes $p$ and $q$ such that wlog $p<q$. The idea is then to use the four number lemma in order to show that $x$ and $y$ have unique factorization, and hence $a$ too has unique factorization, arriving to a contradiction. 
References: 
1) Paul Erdös, János Suranyi, Topics in the Theory of Numbers, Springer, 2003.
2) Steven Weintraub, Factorization: Unique and Otherwise, AK Peters, 2008.
3) David Pengelley, Fred Richman, Did Euclid Need the Euclidean Algorithm to
Prove Unique Factorization?, The American Mathematical Monthly, 113(3), 196-205. Available online here.
4) Lászlo Kalmar, On the Fundamental Theorem of Arithmetic, Matematikai és Fizikai Lapok, (43) 1936, 27-45. (In Hungarian).  
5) W. G. Dubuque, https://math.stackexchange.com/a/322046