Infinite quantity of primes of the form $4k+1$ I need to prove that there are infinitely many primes of the form $4k+1$.
I have proved that $-1$ is not a quadratic residue modulo $4k-1$ and is a quadratic residue modulo $4k+1$. Thus I need to prove that there are infinitely many primes of the form $b^2+1,\ b\in\mathbb{N}$. The standart tehnique "multiply all of them and add 1" doesn't work here. Can anybody help?
 A: Hint: Show that for any finite collection $\{p_1,p_2,\dots,p_n\}$ of primes of the form $4k+1$, there is a prime $p$ of the form $4k+1$ which is not in the collection.  To do this, let 
$$N=(2p_1p_2\cdots p_n)^2+1.$$
Note that $N$ is odd and greater than $1$. So $N$ has an odd prime divisor $p$.
(a) Show that $p$ is not equal to any of the $p_i$.
(b) Note that the congruence $x^2\equiv -1\pmod{p}$ has a solution, namely $2p_1\cdots p_n$.
(c) Conclude that $p$ is of the form $4k+1$.
Remark: A number of similar results can be obtained by using tools from the theory of quadratic residues. For example, one can use arguments of the same general character as the one sketched above to show that there are infinitely many primes of the form $6k+1$, $8k+3$, $8k+5$, $8k+7$, and $10k+9$. 
If $a$ is positive, and $\gcd(a,b)=1$, there are in fact infinitely many primes of the form $ak+b$. This is a much deeper result, called Dirichlet's Theorem on primes in arithmetic progressions.
A: Hint: What you have proven is that if $p\neq 2$, and $p|(b^2+1)$, then $p=4k+1$, since $b^2\equiv -1 \pmod{p}$.  Suppose there were finitely many primes of the form $4k+1$, say $p_1,\cdots, p_k$.  Then consider $$\left(2\prod_{i=1}^k p_k\right)^2+1.$$ What can we say about its prime divisors?  How does that imply there are infinitely many primes of the form $4k+1$?
