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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.

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  • $\begingroup$ You don't have to "sign" your name; it's signed already by your user name. Also, you never stated the theorem that was being proven, did you notice that? $\endgroup$ – Arturo Magidin Mar 3 '11 at 19:10
  • $\begingroup$ "this would imply that $Q$ would also be of this form, which is a contradiction." I don't understand why $Q$ being of this form turns into a contradiction. $\endgroup$ – Valentino Dec 2 '15 at 22:44
  • $\begingroup$ Is this due to the division theorem? $\endgroup$ – Valentino Dec 2 '15 at 22:55
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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.

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  • $\begingroup$ very good observation. That was my typo. $$p_0 = 3$$ was actually true. Thank you. $\endgroup$ – Chan Feb 10 '11 at 15:52
  • $\begingroup$ Please help me by stating the logical reason behind not having any effect on the representation of $Q$ by ignoring $p_0$. Also, in general can we ignore any term(s) while having the same meaning for $Q$. I mean that do we need consecutive values $p_i p_{i+1} p_{i+2} \cdots p_{n-k}$ for $0\le k\lt i\lt n$ & $i,k,n \in \mathbb {Z+}$; or any number of intervening terms can be ignored, as in $p_2p_3p_6p_9\cdots p_{n-3}p_{n}$. $\endgroup$ – jiten Jan 27 '18 at 16:15
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    $\begingroup$ @jiten: if you include $p_0=3$ in the product, the number $Q$ is certainly divisible by $3$. Thus, when you argue that a prime $q$ of the form $4n+3$ divides $Q$ you cannot exclude that actually $q=3$ and therefore you cannot conclude that $q$ is a new prime not already listed. $\endgroup$ – Andrea Mori Jan 28 '18 at 1:30
  • $\begingroup$ Do we see representation by the view of need then, and the feasibility of a representation is just a way to to state an axiom. I mean for $4k+3$ basic fact is under modulo $4$, the only alternate odd prime is $4k+1$ which can't be constructed by a number of the form $4k+3$, and vice-versa. Further explaining why called it an axiom, take odd primes of form $8k+7$, here $p_1=3,p_2=5,p_3=7 \cdots$, but multiplication of $8k+3,8k+5$(can include $8k+1$ also) classes leads to $8k+7$ class. So, in this case, 'elementary' proof for $4k+3$ wouldn't work. Hence, new $Q$ representation doesn't matter. $\endgroup$ – jiten Jan 28 '18 at 9:32
  • $\begingroup$ I mean that if the elementary proof for $4k+3$ works, then need to remove from the left end, else if need to remove from middle (like for $8k+7$ the $p_3$ term), then anew type of proof is needed. So, it is an axiom that $4k+3$ has infinite number of primes, and the representation is just a way to state that. In effect, the feasibility of elementary proof is the key, and based on that have left edge removal case only. $\endgroup$ – jiten Jan 28 '18 at 10:14
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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:

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

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

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