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 ?

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 $4k + 1$.
