Determine the minimum possible amount of distinct prime divisors of $19^{4n}+4$ Determine the minimum possible amount of distinct prime divisors of $19^{4n}+4$, for a positive integer $n$

My works:

By the Sophie Germain factorization, we have
$$19^{4n}+4=(19^{2n}-2\cdot 19^n+2)(19^{2n}+2\cdot 19^n+2)$$

this problem answer is $3$,can you help?
$19^{4n}+4$ has at least 3 prime factors:

Write $19^{4n}+4=xy$ using your factorization. Observe that $x,y$ are relatively prime, since they are both odd and their difference is $4\cdot 19^n$. Moreover, modulo 5 we have $19^{4n}=(-1)^{4n}=1$, so $5\ | \ 19^{4n}+4$. Since neither factor is a power of 5 (see below), the result follows.

Taking $n=1$ yields $19^4+4=5^2\cdot 13\cdot 401$ has 3 distinct prime factors, attaining the minimum.

Silly short proof that neither factor is a power of 5:

Both factors are one more than a perfect square, so by Mihailescu's Theorem they are not powers of 5.
$19^{4n} + 4 = (19^{4n} - 1) + 5 = 0 \pmod 5$