16
$\begingroup$

This is the proof from the book:

Theorem. There are infinitely many primes of the form $4n+3$.

Lemma. If $a$ and $b$ are integers, both of the form $4n + 1$, then the product $ab$ is also in this form.

Proof of Theorem: Let assume that there are only a finite number of primes of the form $4n + 3$, say $p_0, p_1, p_2, \ldots, p_r.$
Let $Q = 4p_1p_2p_3\cdots p_r + 3.$
Then there is at least one prime in the factorization of $Q$ of the form $4n + 3$. Otherwise, all of these primes would be of the form $4n + 1$, and by the Lemma above, this would imply that $Q$ would also be of this form, which is a contradiction. However, none of the prime $p_0, p_1,\ldots, p_n$ divides $Q$. The prime $3$ does not divide $Q$, for if $3|Q$ then $3|(Q-3) = 4p_1p_2p_3\cdots p_r,$ which is a contradiction. Likewise, none of the primes $p_j$ can divides $Q$, because $p_j | Q$ implies $p_j | ( Q - 4p_1p_2\cdots p_r ) = 3$, which is absurd. Hence, there are infinitely many primes of the form $4n +3$. END

From "however, none of the prime ...." to the end, I totally lost!

My questions:

  • Is the author assuming $Q$ is prime or is not?
  • Why none of the primes $p_0, p_1,\ldots, p_r$ divide $Q$? Based on what argument?

Can anyone share me a better proof?

Thanks.

  • 0
    Is this due to the division theorem?2015-12-02

2 Answers 2

11

This is an adaption of Euclid's classical proof of the infinitude of prime. Suppose that $p_1,...,p_t$ are all the primes and consider the number $N=p_1\cdots p_t+1$. The number $N$ must be divisible by some prime (possibly itself, but this is irrelevant for the argument) but since noone of the $p_i$ divides $N$, this gives a contradiction.

The proof you report is similar in concept but is adapted to show that this "extra prime" obtained by looking at divisors of a suitably constructed auxiliary number (the $Q$ in the proof) is actually of the form $4k+3$.

I believe that a slight correction in the proof is in order: namely, take $p_0=3$. The important technical point is that you DON'T include $p_0=3$ in the product defining $Q$. Thus, you can show that none of the $p_i$ (INCLUDING $p_0$) divides $Q$ and you're done by the Lemma.

  • 0
    So, yo mean that for co-prime pairs it works, but what about the representation being tweaked- and its being inconsequential.2018-01-28
9

That part of the proof is simply a variant of Euclid's classical method for producing a new prime. Instead of $\ 1+ \color{#c00}p_{\phantom 1}\!\!\, p_1\cdots p_n\ $ it uses $\ \color{#c00}p+ p_0\cdots p_n,\: $ where $\ (p,\ p_k) = 1\ $ (above $\: p=3,\ p_0 = 4$).
It is easy to verify that this newly constructed integer is coprime to all the prior $\: p $'s, namely:

$$\begin{align} (p,\ \ \: p+p_0\cdots p_n)& = (p,\ p_0\cdots p_n) = 1\ \ {\rm via} \ \ (p,\ p_k) = 1\\[.2em] (p_k,\ p+ p_0\cdots p_n)& = (p_k,\ p)\ =\ 1 \end{align}\quad$$

Essentially this proof relies on the fact that $\ (pq,\ p+q) = 1\! \iff\! (p,\ q) = 1.\ $ Hence to produce a number coprime to $\ n\ $ we can simply sum the factors $\ p,q\ $ from any coprime splitting $\ n = pq.\ $ Euclid's classic proof uses the trivial splitting where $\ q = 1\ $ (and $\:p\:$ is a product of given primes). Ribenboim credits this splitting-form generalization of Euclid's proof to Stieltjes (1890). A slight generalization yields a "coprime" version of Dirichlet's result on primes in arithmetic progression, namely $\,(a,b,c)=1\,\Rightarrow\,(an+b,c) = 1\ $ for some $\,n.$

For a handful of proofs of said gcd property see my post here.