How would you disprove the following statement.. There are an infinite number of primes of the form 4N-1 and a finite number of primes of the form 4N+1.
How would you go about disproving this statement ?
 A: Suppose $p_1, p_2, ... ,p_N$ are all primes in the form $4k +1$. 
$M=(2p_1p_2 ... p_N)^2 + 1$. $M$ is not divisible by $2$ or any number in $p_1, p_2, ..., p_N$. Since none of the factors of $M$ are in the form $4k+1$,  they must be in the form $4k + 3$.
If $p$ is a factor of $M$, then $(2p_1p_2 ... p_N)^2 + 1 = 0 \mod p$, or $(2p_1p_2 ... p_N)^2 = -1 \mod p$.
Quadratic residue tells us $x^2 = -1 \mod p \iff p = 1 \mod 4$, but this is a contradiction because based on our earlier statement $p$ must be in the form $4k + 3$. Therefore $p$ must be in the form $4k +1$ and not part of $p_1, p_2, ... ,p_N$, so it's larger than $p_N$. From this we see there are infinite primes in the form $4k + 1$.
A: One can prove that there are infinitely many primes of the form $4k+1$ by using a variant of the "Euclid" argument. We will need the following useful lemma.
Lemma: Let $m$ be a positive integer. Then any prime divisor of $4m^2+1$ is of the form $4k+1$. 
Proof: Suppose to the contrary that the prime $p$ of the form $4k+3$ divides $4m^2+1$. Note that $\frac{p-1}{2}$ is odd.
Since $p$ divides $4m^2+1$, we have  $(2m)^2\equiv -1\pmod{p}$. Then 
$$(2m)^{p-1}=((2m)^2)^{(p-1)/2}\equiv (-1)^{(p-1)/2}\equiv -1\pmod{p}.$$
This is impossible, since by Fermat's Theorem $(2m)^{p-1}\equiv 1\pmod{p}$.

The rest is as usual. Suppose to the contrary that there are finitely many primes of the form $4k+1$, say $p_1,\dots,p_n$. Let 
$$N=4(p_1p_2\cdots p_n)^2+1.$$
Then $N$ has a prime divisor $p$, and by the Lemma $p$ cannot be of the form $4k+3$. Thus $p$ is of the form $4k+1$, and it is easy to see that $p$ cannot be any of $p_1,p_2,\dots,p_n$.    
