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I found the proof following about infinite prime numbers of form $p = 4m+3$ I tried to find it somewhere on mathstack and I found identical proof but not answer to my question.

Assume that number of prime numbers in form $4m+3$ is finite, then consider product

$$M_k = (4 \cdot 3 \cdot 5\cdot...\cdot p_k) -1$$

We have that $M_k \equiv3(mod\;4)$ So there has to be some prime factor $p = 4m+3$ which divides $M_k$ and it's bigger than $p_k$. Contradiction.

And my question is: why it has to be bigger than $p_k$ ? I tried to rewrite $\frac{M_k}{p}$ but I wasn't able to prove that $p > p_k$ could you please give me a hand with telling me why ?

Bill Dubuque
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Lucian
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    $p$ can't be any of the $p_i$ (the remainder is $1$ after division fo any $p_i$ ), so it has to be bigger than all of them , in particular bigger than $p_k$. – Dietrich Burde Mar 28 '21 at 18:24
  • The proof is explained at length by the answers in the linked dupe. Please read those and pose questions there if any questions still remain. – Bill Dubuque Mar 28 '21 at 18:27
  • As in Euclid's classical proof, the point of adding $\pm 1$ to the product of the prior found primes (of sought form) is that it creates a number $M_k$ that is coprime to all of them (so it has different prime factors, including one of ihe sought form). So if the product in $,M_k,$ has all $,4n+3,$ primes up to $,p_k,$ the new-found one must be larger than $,p_k.,$ It's better not to use contradiction, then we get a constructive proof showing that given any finite list of such primes we can find a new one. That's how Euclid's proof was done. – Bill Dubuque Mar 28 '21 at 18:49

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