Establish that there are infinitely many primes of the form $4k+1$.
I was studying primitive roots, and it was recently proven that the odd prime divisors $n^2 +1$ are all of the form of $4k+1$.
A proof in Burton's Elementary Number Theory assumes that there are finitely many primes of the form $4k+1$. The book lets $N = (2p_1...p_n)^2 +1$ ($p_i$ are the primes of form $4k+1$ then proceeds that this number would have a prime of the form $4k+1$ that is not $p_i$ ($i$ between $1$ and $n$) because $p_i$ does not divide 1, thus does not divide $N$.
I was wondering why you would use $N =(2p_1...p_n)^2+1$, doesn't letting $N=(p_1...p_n)^2 +1$ achieve the exact same thing?