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While reading the Disquisitiones Arithmeticae, Gauss shows a nice proof for the proposition before mentioned, also, I think that I have another, but I don't know if it is correct, correct me if i'm wrong. I'm a young self student and I want to learn from my mistakes. Thanks. And also sorry for some mistakes, I'm not an english native.

If $$abcd \cdots$$ were divisible by p, that would mean that $$abcd \cdots = kp.$$ But, if that would be right, that would mean that one of the values $$a,b,c,d \cdots = p $$(Because p is prime and can't be factored as a product of other two numbers and if abcd... is a multiple of p, one of the prime factors would be p).

But that contradicts our first proposition, so $$abcd...$$ can't be divisible by p.

Bill Dubuque
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Alex Pérez Castilla
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    This doesn't follow. It's not true that if $abcd...$ is divisible by $p$ then one of $a,b,c,d, ...$ must be $p$. e.g. $4*4 = 16$ is divisible by $p=2$... – Jair Taylor Apr 20 '21 at 20:21
  • But this is true because 4 = 2 *2, so at least one of the a,b,c,d... is the prime that is divisible. – Alex Pérez Castilla Apr 20 '21 at 20:26
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    Usually this sort of result is proven as a precursor to unique factorization. As you remark, it follows trivially if you know unique factorization. – lulu Apr 20 '21 at 20:33
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    @Alex you aren't listening to JairTaylor's complaint correctly. In $4\times 6 = 24$ despite $4\times 6$ being a multiple of $2$ neither $4$ nor $6$ is equal to $2$. We aren't told that $a,b,c,d,\dots$ are all primes... There is no reason for one of $a,b,c,\dots$ to be equal to $p$... just that one of them will be a multiple of $p$. – JMoravitz Apr 20 '21 at 20:36
  • Your statement that you claimed, "if $abcd\dots = kp$ then one of the values $a,b,c,d,\dots \color{red}{=} p$" is incorrect. It should have read "one of the elements $a,b,c,\dots$ will be divisible by $p$." The symbol $=$ is not the same as the symbol $\mid$. – JMoravitz Apr 20 '21 at 20:39
  • As for proving this, you ought to share your definition of prime you are using. The most useful definition of prime here is that $p$ is a prime iff for all integers $a,b$ the implication "$p\mid ab\implies p\mid a$ or $p\mid b$" holds. The claim you are looking to prove here is merely the contrapositive extended to an arbitrary number of terms in the product. – JMoravitz Apr 20 '21 at 20:41
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    Yes, you are right @JMoravitz , I appreciate a lot your comments, when I was referring that one of the a,b,c,d = p i was thinking that a,b,c,d were in lower terms eg: 16= 2^4 so a,b,c,d would be 2,2,2,2. Again, thanks for your comments, I've learned a lot for next proofs! – Alex Pérez Castilla Apr 20 '21 at 21:04

2 Answers2

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it is true that if $$ p = abc\cdots z $$ is prime and all the factors are prime then one of the factors is $p$ itself.

How you prove that depends on what you have already proved. It follows easily from the fundamental theorem of arithmetic, which is essentially what you are using in your argument.

But it is usually proved using other arguments and used as an ingredient in the proof of the fundamental theorem.

Where does it appear in Gauss? After the FTA or on the way to it?

Ethan Bolker
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That begs the principle, since you implicitly assume FTA = fundamental theorem of arithmetic. i.e. existence and uniqueness of factorizations into irreducibles. FTA implies Euclid's Lemma, i.e. irreducibles are prime, i.e. if an irreducible divides a product it divides some factor. Conversely, Euclid's Lemma easily yields the uniqueness of factorizations into irreducibles (and this is one of the most common ways to prove it, after proving Euclid using Bezout's theorem for the gcd).

But a proof of the equivalence of these two statements is not a proof that they hold true in some particular number system. There are simple counterexamples showing how they can fail, e.g. the Hilbert numbers $\,\Bbb H = 1+4\:\!\Bbb N = \{1,5,9,13,\ldots\},\, $ the naturals $\equiv 1\pmod{\!4}.\,$ Here there exist nonunique factorizations into irreducibles, e.g. $\, 9\cdot 49 = 21^2.\,$ Thus, by above said equivalence, there also exist counterexamples to Euclid, here $\, 21\mid 9\cdot 49\,$ but $\, 21\nmid 9,49.\,$ See here for more.

Bill Dubuque
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