Nothing I throw at this will stick, Eulers theorem, Fermats theorem feels useless when I apply them on large numbers like this. I'm interested in solving $x^{a}\equiv b \ (\text{mod} \ p),$ where $a=2019$, $b=3571700849900719233$ and $p=2^{64}+13.$
This is what I've done so far:
$$x^{a}\equiv b \ (\text{mod} \ p)\Leftrightarrow x^{a}=b+np, \ \text{for some} \ n\in \mathbb{N}.$$
This means that $\text{gcd}(x,p)=1$ so
$$x^{\phi(p)}=x^{\phi(p-1)}=x^{2^{64}+12}\equiv 1 \ (\text{mod} \ 2^{64}+13),$$
which is exactly Fermats theorem. I also realise that $$x^{2^{64}+12}=x^{2^{64}}x^{2^3}x^{2^2}=(x^{2})^{69}\equiv 1^{69} \ (\text{mod} \ 2^{64}+13),$$
which is equivalent to
$$x^{2}\equiv 1 \ (\text{mod} \ 2^{64}+13).$$
Which feels very wrong and still very hard to proceed. How are these huge numbers treated?
Hint: $p = 2^{64}+13$ is prime, and $2019$ is coprime to $p-1$. If $2019\; \alpha \equiv 1 \bmod (p-1)$, then your answer will be $b^\alpha \bmod p$.