Question #272611

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

1
Expert's answer
2021-12-13T14:00:18-0500

n3(mod4),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 1mod4.1 \bmod 4.

Then prove n is also equivalent to 1mod41 \bmod 4 , which is a contradiction (note (4k+1)(4m+1)=4(km+k+m)+1).(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 {p1,p2,,pn}.\left\{p_{1}, p_{2}, \ldots, p_{n}\right\}. But then

(p1p2pn)2+2=3(mod4)\left(p_{1} p_{2} \cdots p_{n}\right)^{2}+2=3 \quad(\bmod 4)

by (1) has a prime factor pn+1p_{n+1} equivalent to 3mod43 \bmod 4 .

By definition of {p1,p2,,pn}\left\{p_{1}, p_{2}, \ldots, p_{n}\right\} , we get pn+1{p1,p2,,pn},p_{n+1} \in\left\{p_{1}, p_{2}, \ldots, p_{n}\right\},

Thus pn+1(p1p2pn)2+2pn+12p_{n+1}\left|\left(p_{1} p_{2} \cdots p_{n}\right)^{2}+2 \Longrightarrow p_{n+1}\right| 2

Contradiction.


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