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I'm trying to prove that factorization into primes is unique using mathematical induction. Here is my attempt and I would like to know if my reasoning is valid (and advice for any improvements if possible). Thank you.

Proof: Let $A(n)$ be the statement 'Natural number $n$ is unique in prime factorization.'

Initial step: case n = 1,2 is trivial.

Inductive step: Suppose $A(1), A(2), ... A(k)$ are all true for some $k$. If $k+1$ is a prime, it is unique. If $k+1$ is not a prime, it can be written as a product of two natural numbers both smaller than $k+1$, let's say $ k+1 = ab$. Since $a,b$ is both less than $k + 1,$ $A(a), A(b)$ is true so $a$ and $b$ can be written as a unique factorization of prime as follows: $a = p_1^{a_1}p_2^{a_2}...p_i^{a_i}$ and $b = q_1^{b_1}q_2^{b_2}...q_i^{b_i}$. So $k + 1 = ab = p_1^{a_1}q_1^{b_1}p_2^{a_2}q_2^{b_2}...p_i^{a_i}q_i^{b_i}.$ If $p_k = q_k$ then $p_k^{a_k}q_k^{b_k}$ can be written as $p_k^{a_k + b_k}$. Since $k + 1$ is in form $p_1^{e_1}p_2^{e_2}...p_i^{e_i}$, base are unique since they are prime and exponents are unique since they are natural numbers. Therefore all natural numbers can be expressed in unique prime factors by induction.

Mardia
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    How do you know that $k+1$ can't be written as $cd$ where $c,d$ have unique factorizations that use different primes than $a,b$ do? – lulu Aug 14 '20 at 19:37
  • @lulu I assumed using inductive step. – Mardia Aug 14 '20 at 19:39
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    You can't. The inductive hypothesis only applies to numbers $≤k$. – lulu Aug 14 '20 at 19:39
  • but isn't $k+1 > a$ and $k + 1 > b$? – Mardia Aug 14 '20 at 19:41
  • So what? You applied the hypothesis to $a,b$ but I say there's a different factorization $k+1=cd$ that has nothing to do with $a,b$. You haven't said anything to prove me wrong. – lulu Aug 14 '20 at 19:42
  • I thought whatever $cd$ I choose, it's less then $k+1$ so it would all fit in inductive step. – Mardia Aug 14 '20 at 19:45
  • It’s worth seeing how your proof fails when there is no unique factorization. Let $R$ be the ring of numbers of the form $a+b\sqrt{-5},$ where $a,b$ are integers, and with norm $a^2+5b^2.$ your proof works for $R,$ by induction on the norm, but $6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5}).$ – Thomas Andrews Aug 14 '20 at 19:45
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    A good test for supposed proofs of unique factorization is to look at the Hilbert numbers. These are just the integers of the form $4n+1$. Note: you can multiply two of these two get a third. There are primes, i.e. Hilbert Numbers that can not be written as a non-trivial product of other Hilbert numbers. Thus, $5$ is a prime but so is $21$ since neither $3$ nor $7$ are Hilbert numbers. Your argument works in this context, but it shouldn't because unique factorization is false here. $21, 209, 33$ and $133$ are all primes in this sense and $21\times 209=33\times 133$. – lulu Aug 14 '20 at 19:47
  • @lulu I am kind of new to formal math proof so sorry if I am asking too much... I tried to show that all natural number can be written in unique factors of prime. So I guess my proof is wrong but if it was correct should it also be true on the hilbert number system? – Mardia Aug 14 '20 at 20:26
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    My point was that your logic applied to that case as well, and since the theorem is false in that case there must be a flaw in your logic. But I have already said what the flaw is. You start by declaring $k+1=ab$ and then you get to work on $a,b$ but if I started with $k+1=cd$ and got to work on $c,d$ there is no guarantee that we get to the same factorization. I am afraid the problem is a lot harder than that. A lot harder. – lulu Aug 14 '20 at 20:33
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    You should be aware that the proof of Unique Factorization that appears in Euclid is pretty much the proof we use today. It's a brilliant argument and we haven't found a whole lot of ways to improve on it. That said, there are "elementary" proofs using induction...a standard one can be found, e.g. here – lulu Aug 14 '20 at 20:37

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The ingredient you're missing is that if $k+1=\prod_{i\in S}p_i^{A_i}=\prod_{j\in T}q_j^{B_j}$ then$$p_i|k+1=\prod_jq_j^{B_j}\implies\exists j\in T(p_1|q_j)$$where $\implies$ needs a proof by induction on $m:=\sum_jB_j$ of the fact that, if a product of $m$ integers is divisible by a prime $p$, at least one factor is. Hence if $k+1$ has multiple prime factorizations so does $(k+1)/p\le k$, contradicting the null hypothesis.

In case the proof of this theorem about $m$ products is unfamiliar, we proceed as follows:

  • $m=1$ is trivial;
  • We also include $m=2$ in the base case, proven below;
  • If it works for $m=\ell$, a product of $\ell+1$ factors is a product of $2$ factors, the first $\ell$ & the last one, so $m=2$ completes the inductive step.

Soo $m=2$ is the hard part. If $p=uv$ but $p\nmid u$, there are integers $x,\,y$ with $xp+yu=1$ by Bézout's lemma. So $p|xpv+yuv=p$ (because $p|uv$).

J.G.
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