Suppose that S is a finite set of odd primes, each of the form 4k+3 for some natural k. Let z be the product of the numbers in S, plus 3. Would this be a false statement about the prime factors of z?
We can't tell whether they include another 4n+3 prime -- it depends on S.
I have been trying to understand Gaussian but the "plus 3" in the z is confusing me.
The statement is true. If, for example, we take $S=\{3\}$ then $z=6$ and $z$ has no prime factor of the form $4k+3$ other than $3$, which is in $S$. On the other hand, taking $S=\{11\}$ we get $z=14$ which is divisible by $7$.
It is true that $z$ must be divisible by a prime not in S.
To see this note that $$z=\prod_{p_i\in S}\; p_i+3$$ is relatively prime to every $p_i$ except $3$ (if $3$ happens to be one of the $p_i$). Thus the only way $z$ might only have prime factors from S is if $3\in S$ and $z$ is a power of $3$. But, $3^n-3$ is even, hence could not be the product of the primes in S.