Alternative proof to the proposition: 'If a,b,c,d,... are not divisible by p, for p a prime, abcd... also aren't divisible by p' 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.
 A: 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?
A: That begs the principle, since you are implicitly assuming the fundamental theorem of arithmetic. i.e. existence and uniqueness of factorizations into irreducibles ("primes"). That immediately implies Euclid's Lemma (if a prime 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\} = $ naturals with remainder $1$ when divided by $4$. Here there exist nonunique factorizations into irreducibles, e.g. $\, 9\cdot 49 = 21^2,\,$ hence, by above, also counterexamples to Euclid, here $\, 21\mid 9\cdot 49\,$ but $\, 21\nmid 9,49.\,$ See here for more on this.
