= Solution
If a <prime divisor> $p$ of $(n!)^2+1$ satisfied $p\le n$, it would divide $n!$ and therefore divide the difference $(n!)^2+1-(n!)^2=1$, impossible. Hence $p>n$.
Every <integer> $(n!)^2+1>1$ has a <prime divisor>, and (ii) makes each of its <prime divisors> congruent to 1 modulo 4. Taking arbitrarily large $n$ produces such <primes> beyond every bound. Therefore
$$
\boxed{\text{There are infinitely many primes congruent to1 modulo4}.}
$$
Equivalently, if their list were finite, choosing $n$ at least its largest member would produce a new member larger than that supposed maximum.
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