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There are infinitely many primes of the form $4n+3$. Proof: Define $q$ by $q=2^2.3.5...p-1$. Then $q$ is of the form $4n+3$, and is not divisible by any of the primes up to $p$. It cannot be product of primes $4n+1$ only, since the product of two numbers of this form is of same form; and therefore it is divisible by a prime $4n+3$, greater than $p$.

I don't understand how "It cannot be product of primes $4n+1$ only, since the product of two numbers of this form is of same form"have conclusion "therefore it is divisible by prime greater than $p$" ? Also how $4n+1$ came out from thin air? Mine version of explanation will exclude $4n+1$ since you can literally say $4n+3$ is prime the nothing to prove else there are prime not included in that weird $q$ so using fundamental theorem of arithmetic there is one more prime number. Also how $q$ is derived? My may is you first have multiple of prime then you need to force this to be of form $4n+3$ by $4(d+q)+3$ where $q$ is $-1$ and $d$ is multiple of prime number finite amount I think it should be prime number of form $4n+3$ in $d$ which is missing from textbook definition. I hope I get absolute correct and detailed explanation. Thanks for attention!

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    $\begingroup$ A product of numbers of the form $4n+1$ also has this form. This is the key for this proof. This guarantees that there must be a prime factor of the form $4n+3$ $\endgroup$ – Peter Sep 14 '20 at 12:18
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    $\begingroup$ A prime $>2$ is either in the form of $4n+1$ or $4n+3$. Suppose $q$ has no prime factor of the form $4n+3$. Then all prime factors of $q$ is of the form $4n+1$. $\endgroup$ – player3236 Sep 14 '20 at 12:23
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    $\begingroup$ OK, $q$ is odd, so it only has odd prime factors. But every odd prime number (in fact every odd number) is either of the form $4n+1$ or of the form $4n+3$. If all prime factors were of the form $4n+1$ , $q$ would be of the form $4n+1$ as well which is not the case. $\endgroup$ – Peter Sep 14 '20 at 12:23
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    $\begingroup$ We can reduce $4n+5,4n+7$ etc. to the case $4n+1$ or $4n+3$ if we choose $n$ suitable. And in fact, Euklid's idea is imitated. $q$ cannot have a prime factor less than or equal to $p$. $\endgroup$ – Peter Sep 14 '20 at 12:32
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    $\begingroup$ If we multiply two numbers of the form $4n+1$ , we have $$(4a+1)\cdot(4b+1)=16ab+4a+4b+1=4(4ab+a+b)+1$$ so we again get a number of the form $4n+1$ $\endgroup$ – Peter Sep 14 '20 at 12:40
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Summarizing the comments, note that all odd numbers and therefore all primes greater than $2$ can be expressed as $4n+1$ or $4n+3$. A number of the form $4n+5=4(n+1)+1$ is of the first form.

Second, the product of numbers of the form $4n+1$ is again of the form $4n+1$. We can see that by writing $(4n+1)(4m+1)=16nm+4(n+m)+1$

Third, assume that there are a finite number of primes of the form $4n+3$. Among them are $3,7,11,19 \ldots$. Let the greatest of them be $p$. Form the number $N=2^2\cdot 3 \cdot 5 \cdot 7 \ldots p-1$ where the factors are all the primes up to $p$. $N$ is of the form $4n+3$ and is not divisible by any of the primes up to $p$. By our hypothesis it must be a product of primes of the form $4n+1$ because all the primes of form $4n+3$ are less than or equal to $p$. Since the product of numbers of the form $4n+1$ is again of the form $4n+1$ we have a contradiction.

Therefore there are infinitely many primes of the form $4n+3$.

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