 Solution

Suppose all primes congruent to $3$ modulo eight are $p_1,\dots ,p_r$, and put $N=p_1\cdots p_r$, an odd integer. The integer $N^2+2$ is odd and congruent to three modulo eight. If a prime $q$ divides it, then $N^2\equiv -2(\mathrm{mod}q)$ and $q\nmid N$, since a common divisor of $N$ and $N^2+2$ would divide two. Thus $(-2/q)=1$, so part (v) forces $q\equiv 1$ or $3(\mathrm{mod}8)$.

If all prime factors congruent to three had even multiplicity, their product, including the factors congruent to one, would be one modulo eight. Since $N^2+2$ is three modulo eight, at least one prime factor is congruent to three and occurs to odd multiplicity. But it is none of the $p_i$, since those divide $N$. This contradiction proves infinitely many primes are congruent to three modulo eight.