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Update: If this question is of interest, you can also click here.

Update: Using Bill Dubuque's lemma and logic proving Euclid's lemma, we can supply an elementary proof.

To get a contradiction, assume than $p \mid a b$.

Let $S = \{n \in \Bbb N \, | \, p \mid nb \}$. Then $p \in S$ and $a \in S$. Moreover, $S$ is closed under subtraction.

Let $d = \text{min(}S\text{)}$. By the lemma, $d \mid p$, so $d = 1$ or $d = p$.

If $d = 1$, since $d \in S$, it must follow that $p \mid (1 \times b)$, which is absurd since $b \lt p$.

By the lemma, $d \mid a$, so if $d = p$ then $p \mid a$, which is absurd since $a \lt p$.

I've been motivated (see this) to prove the following result using only elementary techniques.

Let $p$ be a prime greater than $2$.

Let $1 \lt a \lt p$

Let $1 \lt b \lt p$

Then

$$\tag 1 p\nmid a b$$

I think that this is as simple as first showing that

$$ \text{For every integer n } \ge 1 \text{ such that } p\nmid n, \; \; p\nmid na$$

and ironing out some details.

Using only the 'first page' of elementary theory of the natural numbers/integers (for example Euclidean division, the construction of $\Bbb Z$, the existence of prime factorizations and that modular arithmetic is well-defined), can this approach work for proving $\text{(1)}$?

Besides answering yes in the comments, a proof would be appreciated (this elementary approach can be exhausting).

CopyPasteIt
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  • From the definition of prime number, I think it is easy to show that $p\mid ab\implies p\mid a\lor p\mid b$; so the contrapositive of this result will establish your desired result (EDIT: I guess proving the result I say here would take a bit more than the definition of primes). As far as your proposed approach goes, I think that it would be more work than needed. – Dave May 21 '19 at 17:09
  • I gave a couple direct proofs by induction in this thread. But it's not clear what you mean by "no number theory", so it's not clear if they suffice. E – Bill Dubuque May 21 '19 at 17:30
  • See also this proof of Euclid's Lemma by inductively proving that a fraction writable with coprime denominators is an integer. There are many variants of these "direct" proofs but they all are essentially "assembly language" compilations of higher-level more conceptual proofs. – Bill Dubuque May 21 '19 at 17:46
  • So, while at first glance they don't appear to "use number theory", that's only because the use has been highly obfuscated by eliminating all the higher-level concepts, e.g. directly inlining inductive proofs of gcd properties by repeatedly subtracting, etc. I elaborate on this in the answers and their links. – Bill Dubuque May 21 '19 at 17:57
  • And another way is to use Gauss's algorithm to show that $a$ is invertible $\bmod p,$ (and the inductive proof yields an algorithm to compute the inverse, essentially a special case of the (extended) Euclidean algorithm). – Bill Dubuque May 21 '19 at 18:21
  • Thanks @BillDubuque for all the links and comments. I will review them and try to get out a proof. As for the amount of number theory, I thought it could all be developed ab initio. But of course you can always just state up front what you want to use as a starting point. – CopyPasteIt May 21 '19 at 20:41
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    Re: your edit, all 4 proofs I linked seem to satisfy your "front page" criterion. But if you are interested in mastering number theory then you shouldn't seek these arithmetical assembly language versions. Rather, you should seek to understand their higher-level conceptual counterparts (which generalize to other number rings, e.g. Euclidean or gcd domains). What is the reason you are interested in peeking at the assembly language? – Bill Dubuque May 21 '19 at 21:23
  • @BillDubuque I'm imagining this work being fed to an $\text{AI MATH BOT}$ to see what it can do. I know, too much time on my hands. From elementary foundation wanted to also show $\quad$ $d \mid n$ and $n = ab$ then we can tit-for-tat factor $d$. $\quad$ $d = xy$ where $x \mid a$ and $y \mid b$. $\quad$ $\frac{a : b}{x : y}$ – CopyPasteIt May 21 '19 at 21:37
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    For that see Euler's Four Number Theorem and related properties described there. – Bill Dubuque May 21 '19 at 22:18
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    The proof you posted is just the contrapositive form of the proof I gave above the Lemma you linked to. – Bill Dubuque May 22 '19 at 17:58
  • @bill yes. I know - almost copied your proof using the lemma. This question came up on my worksheets so it was the one I wanted answered. I am hoping to revisit https://math.stackexchange.com/q/3235873/432081 without relying on this. Just skating around Euclid's lemma for now... – CopyPasteIt May 22 '19 at 18:06
  • Ok, wasn't sure if you realized that. It's not clear what you are aiming at. – Bill Dubuque May 22 '19 at 18:13

3 Answers3

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You need some more than the definition of prime to prove this. The reason is that in other rings (systems of 'numbers' [they can actually be numbers, polynomials or any 'thing' that you can sum and multiply] with sum and product) there are elements that can be only divided (essentially) by $1$ and themselves, but there are pairs of this 'numbers' $a,b$ such that $p\mid ab$ but $p$ does not divide $a$ or $b$. A well-known example is $2\cdot3=(\sqrt {-5}+1)(-\sqrt {-5}+1)$.

The best-trodden path for natural numbers is through Euclid's algorithm and Bezout's identity. It is not difficult to read, but kind of long to post here and easy to find.

ajotatxe
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    I am not concerned with abstract ring theory. I am looking only at the natural numbers. I know about Euclid's algorithm and Bezout's identity but trying to logic-boot this up from less theory. – CopyPasteIt May 21 '19 at 20:47
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If p divides ab, then it divides a or b (I'm not sure if you covered this; you can try to prove this as an exercise otherwise). This implies that it is less than or equal to a or b (since 1 is less than a and b), which contradicts the fact that a is less than p and b is less than p.

availableusername
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Well,that's not true Just take one counterexample p=7 (prime greater than 2) Take a=3 b=4 3.4=12 which is not divisible by 7

Prastya Susanto
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