Prime

There are infinite number of prime numbers, such that it can be expressed in a $4k-1$ for, where k is an integer.

First note that every $n$ integer $n\equiv3 \ \mod4$ has a $m$ divisor, where $m\equiv3 \ \mod4$ . (Since all of $n$ 's divisors is odd, and then if each divisor of $n$ makes $1$ remainder when it is divided by $4$ , then $n$ would make $1*1*1*\dots*1$ remainder when it is divided by $4$ .) So suppose there are finite prime numbers which make $3$ remainder ( $\mod4$ ): $p_1, p_2, \dots, p_r$ . Let $N=4p_1p_2p_3\dots p_r-1$ Then $N$ has an $x$ divisor, which can be expressed in a $4l-1$ formula ( $l$ is an integer), but then $x|1$ , which is a contradiction.

There are infinite number of prime numbers, such that it can be expressed in a $4k+1$ for, where k is an integer.

By using congruence it is easy to prove that for every $n$ integer $n^2+1$ has an $x$ divisor which makes $1$ remainder when it is divided by $4$ . So suppose there are finite prime numbers which make $1$ remainder ( $\mod4$ ): $p_1, p_2, \dots, p_s$ . Let $N=4(p_1p_2p_3\dots p_s)^2+1$ Then $N$ has an $x$ divisor, which can be expressed in a $4q+1$ formula ( $q$ is an integer), but then there exist a $p$ prime, where $p\equiv1 \ \mod4$ and $p$ is not part of the $p_1,p_2, \dots, p_s$ sequence, which is a contradiction.