Show that if $n \equiv 3\pmod{4}$, then n has a prime factor $p \equiv 3\pmod{4}$

Question

Show that if $n\equiv 3\pmod{4}$, then $n$ has a prime factor $p\equiv 3\pmod{4}$

My approach:

By definition any composite number can be represented as a product of primes, so let $n=p_1\cdots p_k$. $$p_1\cdots p_k \equiv 3\pmod{4}$$

If there is not such prime number then all primes are congruent to either 1 or 2 mod 4 which can't be possible hence there is no way to get $3\pmod{4}$ by multiplying something 1 or 2 $\pmod{4}$

I think this is very informal, so I am just wondering if there is a way to express my idea with a better mathematical argument

If $n\equiv 3\pmod{4}$, then $n$ is odd and so all its prime factors are odd and congruent to $1$ mod $4$. — user236182, Jul 09 '16 at 07:37
wait why?, how do you conclude b=0? $7 \equiv 3 (mod\text{ } 4)$ — TheMathNoob, Jul 09 '16 at 07:39
Got it, it can't be congruent to $2(mod\text{ } 4)$ hence p is odd — TheMathNoob, Jul 09 '16 at 07:43
It isessential to justify the claim "which can't be possible". Also "any prime number" should be "any integer $n > 1$" (and if $n$ is allowed negative then account for sign). — Bill Dubuque, Jul 09 '16 at 14:40

score 4 · Answer 1 · answered Jul 09 '16 at 07:36

Suppose that $n$ does not have a prime factor $p\equiv 3\bmod 4$. By assumption then all prime factors have to be odd, and congruent $1$ modulo $4$. However, then also the product is congruent $1$ modulo $4$, a contradiction. The computation is as follows (this is where you are too informal, perhaps). If $p=4n+1$ and $q=4m+1$, then $$pq=4(4nm+n+m)+1\equiv 1 \bmod 4.$$

Bill Dubuque · Answer 2 · 2025-01-21T20:55:45.127

The set $\:\!S\:\!$ of integers $\,\equiv \color{#c00}1\pmod {\!4}\,$ is closed under multiplication (by $\:\!\color{#c00}{1^n\!\equiv 1}),\,$ so

$${\rm all}\ \ p_i \in S\ \Rightarrow\ p_1 p_2 \cdots p_k \in S$$

The complement (contrapositive) form of this multiplicative closure of $\,S\,$ is

$$p_1 p_2 \cdots p_k \not \in S\ \Rightarrow\ {\rm some}\ p_i\not\in S\quad$$

hence $\,p_i\not\in S\,\Rightarrow \bmod 4\!:\ p_i\color{#c00}{\not\equiv 1}\,$ so $\,p_i\equiv 3\,$ (by $\,p_i\,$ odd, $ $ by $\ p_i\mid n\,$ odd).

Remark $ $ Such "complementary" views of algebraic closure properties are ubiquitous (e.g. see this list and these answers) so it is well worth learning to efficiently work with them.

Daniel Buck · Answer 3 · 2016-07-24T19:08:10.090

If $n \equiv 3 \pmod{4}$ then $n$ contains an odd amount of prime factors of the form $p \equiv 3 \pmod{4}$.

To see this consider three cases of the product of two primes:

If all primes are of the form $p=4k+1$ so is their product of that form: $$p_1\cdot p_2=(4j+1)(4k+1)=4(4kj+j+k)+1$$ whereas the other two cases to consider are: $$p_1\cdot p_2=(4j+1)(4k+3)=4(4kj+3j+4k)+3$$ $$p_1\cdot p_2=(4j+3)(4k+3)=4(4kj+3(j+k)+2)+1$$ Now induct on these results to show that if $n$ has an odd number of primes of the form $4k+3$ in its factorisation then $n\equiv 3\pmod{4}$; otherwise if $n\ $s prime factorisation consists of any number of $p=4k+1$ primes along with an even number of $q=4j+3$ primes we have $n\equiv 1\pmod{4}$.

Show that if $n \equiv 3\pmod{4}$, then n has a prime factor $p \equiv 3\pmod{4}$

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