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Is to possible to prove the problem with elementary approach as used to prove the case $4n+3$. Most of the proof that proves Infinitude of primes of the form $4n+1$ uses the some theorem from quadratic reciprocity.

So I was curious to know whether this proof can also be done as the same way as of the proof for the case $4n+1$ without using any special result.

I am aware of the proof of this fact available in this site. But I just want the proof in the way this book mentioned.

Any help would be appreciated. Thanks in advance.

math is fun
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    What goes wrong when you just follow the suggested method? It's practically a complete solution... – lulu Jun 05 '20 at 16:21
  • Product of numbers of the form$4n+3$ can be of the form$4n+1$ – math is fun Jun 05 '20 at 16:25
  • I meant, the method suggested by the problem you quote. All you need to note is that $-1$ can't be a square $\pmod p$ if $p=4k+3$. That's a very basic point, nothing nearly as hard as Quadratic Reciprocity. – lulu Jun 05 '20 at 16:26
  • Can you explain it little more? – math is fun Jun 05 '20 at 16:29
  • What I wrote is essentially equivalent to the accepted solution here, you just need to modify it (in a minor way) to fit the exact form in the suggestion. – lulu Jun 05 '20 at 16:30
  • But how do you prove the fact you just mentioned above – math is fun Jun 05 '20 at 16:33
  • Please study the accepted answer in the duplicate I linked to. That answer covers that point. Or you could try to prove it yourself as an (elementary) exercise. Just use Fermat's Little Theorem. – lulu Jun 05 '20 at 16:35
  • But here the integer is not if the form $n^2 +1$ rather it is of the form $n^2 +4$ . So how does it apply in this case? – math is fun Jun 05 '20 at 16:38
  • Please think about it for a second. I think you can sort these details out by yourself with minimal effort. – lulu Jun 05 '20 at 16:40

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The easiest way would be the one highlighted above. The main tool you use to show the infinitude of primes of the form $4n+3$ is the fact that if a number $x$ satisfies: $$x \equiv 3 \pmod{4}$$ then there exists a prime divisor which is $3 \bmod{4}$. This isn't true when you replace $3 \bmod{4}$ by $1 \bmod{4}$.

Thus, we have to use the first supplement of quadratic reciprocity to prove the infinitude of primes which are $1 \bmod{4}$. This is one of the simplest ways we can prove the same.

Haran
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  • But the way writer presented the idea of the proof forced me to think it might be done more elementarily. I know the method used in the case $4n+3$ case wouldn't work here. But I was thinking whether any other elementary method would work. – math is fun Jun 05 '20 at 16:24
  • @mathisfun The only reason why the $4n+3$ case was so elementary was because you could apply that manipulation. The idea of getting a contradiction is based on construction. The whole motivation behind construction is the property you are using. For the $4n+3$ case, we used the fact above. For the $4n+1$ case, the most elementary special property that distinguishes primes of this form from primes of the form $4n+3$ is the first supplement of quadratic reciprocity. Besides, if there was an easier solution, it would have definitely been found and would be popular by now. – Haran Jun 05 '20 at 16:30