Three years, back and forth #2

Since $2015 \equiv -1 \pmod{16}$ and the tetration ${}^42015 = 2015^{2015^{2015^{2015}}}$ is odd, we see that $f(2015) = {}^52015 \equiv -1 \pmod{16}$ . On the other hand, $2015 \equiv 0 \pmod{5}$ , and hence $f(2015) \equiv 0 \pmod{625}$ . Using the Chinese Remainder Theorem, we deduce that $f(2015) \equiv 9375 \pmod{10000}$ .

Since $2018$ is even, it is clear that $f(2018) \equiv 0 \pmod{16}$ . Note now that $2018^4 \equiv 1 \pmod{25}$ , so that $2018^{100} \equiv 1 \pmod{625}$ . Since it is clear that ${}^32018 \equiv 0 \pmod{4}$ , we deduce that ${}^42018 \equiv 1 \pmod{25}$ . On the other hand it is clear that ${}^42018 \equiv 0 \pmod{4}$ . Using the Chinese Remainder Theorem, we see that ${}^42018 \equiv 76 \pmod{100}$ , and hence $f(2018) \equiv 2018^{76} \equiv 401 \pmod{625}$ .Using the Chinese Remainder Theorem again, we have that $f(2018) \equiv 9776 \pmod{10000}$ .

Note that $2021 \equiv 1 \pmod{4}$ and that $2021^4 \equiv 1 \pmod{16}$ . Thus it is clear that ${}^42021 \equiv 1 \pmod{4}$ and hence that $f(2021) \equiv 2021 \equiv 5 \pmod{16}$ . Now note that $2021^5 \equiv 1 \pmod{25}$ , $2021^{25} \equiv 1 \pmod{125}$ and $2021^{125} \equiv 1 \pmod{625}$ . Since $2021^{2021} \equiv 1 \pmod{5}$ , we deduce that ${}^32021 \equiv 2021 \equiv 21 \pmod{25}$ , so that ${}^42021 \equiv 2021^{21} \equiv 46 \pmod{125}$ , and thus $f(2021) \equiv 2021^{46} \equiv 46 \pmod{625}$ . Using the Chinese Remainder Theorem one more time, $f(2021) \equiv 4421 \pmod{10000}$ .

Thus $f(2015) + f(2018) + f(2021) \; \equiv \; 9375 + 9776 + 4421 \; \equiv \; \boxed{3572} \pmod{10000}$