Show that if $n \equiv 3\pmod{4}$, then n has a prime factor $p \equiv 3\pmod{4}$ 
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
 A: 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
$$pq=4(4nm+n+m)+1\equiv 1 \bmod 4.$$ 
A: Hint $\ $ The set $\,S\,$ of integers $\,\equiv 1\pmod 4\,$ are closed under multiplication, therefore
$${\rm all}\ \ p_i \in S\ \Rightarrow\ p_1 p_2 \cdots p_k \in S$$
The complement (contrapositive) form of this multiplicative closure of $\,S\,$ is
$$p_1 p_2 \cdots p_k \not \in S\ \Rightarrow\ {\rm some}\ p_i\not\in S$$
Thus $\,p_i \not\in S,\ $ i.e. $\ p_i\not\equiv 1\pmod 4,\ $ so $\,p_i\equiv 3\pmod 4 \ $ by $\,p_i\,$ odd, $ $ by $\ p_i\mid n\,$ odd.
Remark $\ $ Such "complementary" views of algebraic closure properties are ubiquitous, so it is well worth learning to efficiently work with them.
A: 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 $n\equiv 1\pmod{4}$.
