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So it's a different take on proving there are infinite primes

Given a sequence where any two terms in the sequence are pairwise coprime with each other, how can you prove there are an infinite number of primes combined with the fundamental theorem of arithmetic?

Obviously it follows that if any two of the terms are coprime then their gcd is 1 but I don't see how I can combine this with the fundamental theorem of arithmetic.

Cheers for any help guys

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    A finite sequence or an infinite sequence? – James Arathoon Feb 04 '19 at 15:50
  • Does this really lead to a proof that there are infinite primes, or are you just proving that your initial assumption is a statement consistent with the existence of infinite primes? – James Arathoon Feb 04 '19 at 15:56
  • Well my textbook gives an example of a sequence showing the first few terms are all pairwise coprime, It says assuming every term in the sequence is pairwise coprime how does this combined with the fta lead to a proof of infinite primes? –  Feb 04 '19 at 16:01
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    Seems like circular reasoning to me. How do you prove in the first place that an infinite sequence of pairwise coprime numbers is indeed possible, rather than just assuming it so? – James Arathoon Feb 04 '19 at 16:11
  • Aargh. It's not true that there are infinite primes! Every prime is finite. There are infinitely any primes. The set of primes is infinite. – David C. Ullrich Feb 04 '19 at 20:55
  • Is that aimed at me or James? –  Feb 04 '19 at 21:08
  • Related: https://math.stackexchange.com/questions/50006/different-ways-to-prove-there-are-infinitely-many-primes (for example with Fermat numbers) – Henry Aug 04 '20 at 00:05

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Note: The sequence $\{a_n\}$ defined by $a_n=1$ for all $n$ is a counterexample to this. In what follows, I'll assume that the $a_n$ are $>1$. Easy to generalize to, say $|a_n|>1$ or to something like "infinitely many of the $a_n$ are $>1$ in absolute value.

Let your sequence be $\{a_n\}$ and, for each $n$, define $p_n$ to be the least prime dividing $a_n$.

Your assumption tells us that the $p_n$ are all distinct, since $$p_i=p_j=p\implies p\,|\,\gcd(a_i,a_j)\implies i=j$$

In this way we've produced an infinite sequence of distinct primes.

lulu
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  • Thanks for your answer, it might sound silly but where is the fundamental theorem of arithmetic being used here though? –  Feb 04 '19 at 17:24
  • @BigWilfy Well, I'd say it isn't there. All I need to know is that every natural number ($>1$) is divisible by some prime. It then makes sense to speak of the least prime dividing a given number. If you prefer, you can use unique factorization to write down the prime factorization of each $a_n$ and then select the least prime that appears in that. – lulu Feb 04 '19 at 17:57
  • @BigWilfy Note: I added a note to my post to exclude degenerate cases, such as the sequence ${1,1,1,1,1,\cdots}$. Technical point, but worth noting. – lulu Feb 04 '19 at 17:58
  • Thanks again, So I need to explicitly say what the lowest prime is in the sequence or? sorry I feel a bit lost currently –  Feb 04 '19 at 18:30
  • @BigWilfy Not sure where you are coming up with that...$p_n$ is the least prime dividing $a_n$. Say, for example, $a_n=2^{(2^n)}+1$, the $n^{th}$ Fermat number. It is a well known theorem that these are pairwise relatively prime. In that case $p_1=5$ since $5$ is the least prime dividing $2^2+1$. Using a computer we get $p_4=65537$ and $p_5=641$ so the primes $p_n$ need not be increasing. – lulu Feb 04 '19 at 18:35
  • Can I not just say that for our sequence (an) as any two elements of this sequence are pairwise coprime then this will contain infinitely main primes as each new element in the sequence has a new prime factor (where n is an element of the naturals) Or is this totally off? –  Feb 04 '19 at 19:23
  • @BigWilfy Well, how does that differ from what I wrote? I just took the extra step of explicitly "choosing" one of the primes dividing $a_n$. Phrased differently, my approach is an algorithmic form of yours. Given a specific sequence $a_n$ I give an explicit method for choosing a prime. You could have taken the largest, or any other rule you happen to like. – lulu Feb 04 '19 at 19:26
  • But does this not technically use the fundamental theorem of arithmetic by breaking any number down in the sequence by it's prime factors? Sorry i'm not saying you're wrong btw just trying to understand it fully :) –  Feb 04 '19 at 19:30
  • @BigWilfy Saying that a natural number $>1$ is divisible by a prime is a lot easier to prove than unique factorization. To be sure, you are free to use unique factorization if you want! But I think it's a always good, in a proof, to use the weakest theorems you can. – lulu Feb 04 '19 at 19:35
  • Ah okay thanks, I was just trying to prove it with fta as it said they could be combined together to create another proof of the infinite primes –  Feb 04 '19 at 19:42
  • @BigWilfy Sorry, I wrote something incorrect in my last comment. The main point, however, stands: I expect that my argument, or some minor variation of it, is what is expected here. But of course you haven't proven anything until you actually exhibit a sequence ${a_n}$ with the desired properties. – lulu Feb 04 '19 at 19:49
  • @BigWilfy This old idea goes back to Hurwitze (1891) if not ealrier (e.g. Goldbach's use of Fermat numbers in 1730). Please search before asking common questions that are likely to be asked and answered many times before. – Bill Dubuque Feb 04 '19 at 20:21
  • @BigWilfy You might be interested in this old question. This provides several sequences that fit the bill. Taking $a=2$ gives the Fermat example I mentioned earlier. – lulu Feb 04 '19 at 20:29