= Solution
Suppose there are only finitely many <primes> $p_1,\ldots,p_r\equiv3\pmod8$, and let $N$ be their product. This is odd, so $N^2+2\equiv3\pmod8$. No listed <prime> divides this number, since reduction modulo any $p_i$ gives two. Every <prime> divisor $q$ is odd and does not divide $N$, and satisfies $N^2\equiv-2\pmod q$. Thus $(-2/q)=1$, forcing $q\equiv1$ or $3\pmod8$ by (v).
A product of <prime> factors all congruent to one modulo eight could not be three modulo eight. More precisely, the product of factors congruent to three is three modulo eight exactly when their total exponent is odd. Thus at least one <prime> divisor $q\equiv3\pmod8$ occurs. It is not in the alleged <complete> list, a contradiction. Hence \b[there are infinitely many <primes> congruent to three modulo eight].
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