Answer to Question #272611 in Combinatorics | Number Theory for Prathibha Rose

Question

Prove that n=2 mod4,and when n=p ,where p is a prime and p≡3 mod4

Answer 1

$n \equiv 3(\bmod 4),$ therefore all the prime factors of n are odd.

For contradiction, assume all of the prime factors are equivalent to $1 \bmod 4.$

Then prove n is also equivalent to $1 \bmod 4$ , which is a contradiction (note $(4 k+1)(4 m+1)=4(k m+k+m)+1).$

 Suppose for contradiction there are only finitely many primes of the form 4k+3, and call the set of all of them $\left\{p_{1}, p_{2}, \ldots, p_{n}\right\}.$ But then

$\left(p_{1} p_{2} \cdots p_{n}\right)^{2}+2=3 \quad(\bmod 4)$

by (1) has a prime factor $p_{n+1}$ equivalent to $3 \bmod 4$ .

By definition of $\left\{p_{1}, p_{2}, \ldots, p_{n}\right\}$ , we get $p_{n+1} \in\left\{p_{1}, p_{2}, \ldots, p_{n}\right\},$

Thus $p_{n+1}\left|\left(p_{1} p_{2} \cdots p_{n}\right)^{2}+2 \Longrightarrow p_{n+1}\right| 2$

Contradiction.