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I have to prove two things, but I don't know where to start. Can anyone offer some guidance?

$1.$ Let $n\in\mathbb{N}$. Show that if $n$ is congruent to $3 \pmod 4$, then $n$ has a prime factor which is congruent to $3 \pmod 4$.

$2.$ Show that there is infinitely many prime numbers congruent to $3 \pmod 4$.

UPDATE:

I have trouble understanding the questions. Could someone provide a walkthrough to the proof of both 1) and 2)?

jukka.aalto
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3 Answers3

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1) $n\equiv 3\pmod{4}$, therefore all the prime factors of $n$ are odd. For contradiction, assume all of the prime factors are equivalent to $1$ mod $4$. Then prove $n$ is also equivalent to $1$ mod $4$, which is a contradiction (note $(4k+1)(4m+1)=4(km+k+m)+1$).

2) Suppose for contradiction there are only finitely many primes of the form $4k+3$, and call the set of all of them $\{p_1,p_2,\ldots,p_n\}$. But then $$\left(p_1p_2\cdots p_n\right)^2+2\equiv 3\pmod{4}$$

by (1) has a prime factor $p_{n+1}$ equivalent to $3$ mod $4$. By definition of $\{p_1,p_2,\ldots,p_n\}$, we get $p_{n+1}\in\{p_1,p_2,\ldots, p_n\}$, thus $$p_{n+1}\mid \left(p_1p_2\cdots p_n\right)^2+2\implies p_{n+1}\mid 2,$$ contradiction.

user236182
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  • Why assume the prime factors are equivalent to 1 mod 4? Can you explain more? – jukka.aalto Sep 14 '15 at 16:32
  • @jukka.aalto You prove (1), (2) using a proof by contradiction. If you know every prime factor can't be $\equiv 1\pmod 4$, then this proves at least one of the prime factors must be \equiv 3\pmod$, which is what we need to prove. – user236182 Sep 14 '15 at 17:59
  • Why does the fact that every prime factor to $n$ can't be equivalent to 1 mod 4 mean that at least one of the prime factors is equivalent to 3 mod 4? – jukka.aalto Sep 16 '15 at 07:04
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    @jukka.aalto Every prime factor of $n$ can't be equivalent to $1$ mod $4$, therefore at least one of the prime factors of $n$ is equivalent to $0$ or $2$ or $3$ mod $4$. But $n$ is odd, so every prime factor of $n$ is odd, therefore at least one of the prime factors of $n$ is equivalent to $3$ mod $4$. – user236182 Sep 16 '15 at 13:21
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Hint: For 1, prove the contrapositive: if all prime factors of an odd number $n$ are congruent to 1 modulo 4, then so is $n$. Then extend this to even numbers.

lhf
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For the second exercise a prime number $p>3$ can be written in the form $6n\pm 1$ therefore a number prime can be written as $$p=6n+1$$. You can control the congruence module $4$. Indeed $$p=6n+1\equiv 2+1=3 \pmod 4$$ Edit: the thing works if $$n=4k+1$$

Domenico Vuono
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