There are infinitely many primes whose remainder is 3 when divided by 4.
Proof:If there are only finite number of such primes,suppose they are
\begin{equation}
p_1,p_2,p_3,\cdots,p_n
\end{equation}
Now let's see
\begin{equation}
p_1p_2\cdots p_n
\end{equation}
When $n$ is odd,it is easy to verify that $p_1\cdots p_n\equiv 3\mod 4$,then let's see
\begin{equation}\label{eq:feel}
p_1p_2\cdots p_n+4
\end{equation}
It is obvious that \ref{eq:feel} is a prime of the form $4k+3$(How?)
When $n$ is even,it is easy to verify that $p_1p_2\cdots p_n\equiv 1\mod 4$,then let's see
\begin{equation}\label{eq:scbdsh}
p_1p_2\cdots p_n+2
\end{equation}
It is easy to verify that \ref{eq:scbdsh} is a prime of the form $4k+3$.Whatever the case,we construct a prime number of the form $4k+3$ which equals to none of $p_1p_2\cdots p_n$,this leads to absurdity.So there are infinitely many primes of the form $4k+3$.