8. There is a unique function $f: \mathbb{N} \rightarrow \mathbb{R}$ such that $f(1)>0$ and such that
$$
\sum_{d \mid n} f(d) f\left(\frac{n}{d}\right)=1
$$
for all $n \geq 1$. What is $f\left(2018^{2019}\right)$ ?
Answer:
$\frac{\binom{4038}{2019}^{2}}{2^{8076}}$ OR $\frac{(4038!)^{2}}{(2019!)^{4} \cdot 2^{8076}}$

Fix any prime $p$, and let $a_{n}=f\left(p^{n}\right)$ for $n \geq 0$. Notice that using the relation for $p^{n}$, we obtain
$$
\sum_{i=0}^{n} a_{i} a_{n-i}=1,
$$
which means that if we let $g(x)=\sum_{n \geq 0} a_{n} x^{n}$, then $g(x)^{2}=1+x+x^{2}+\cdots=\frac{1}{1-x}$ as a generating function. Thus $g(x)=(1-x)^{-\frac{1}{2}}$, and this is well-known to have generating function with coefficients intimately related to the generating function for the Catalan numbers. In particular, $a_{n}$ is independent of our choice of $p$.
Now if we define $f_{0}(n)=\prod_{p \mid n} a_{v_{p}(n)}$, then we see that $f=f_{0}$ on the prime powers.
If we define the Dirichlet convolution of two functions $\chi_{1}, \chi_{2}: \mathbb{N} \rightarrow \mathbb{R}$ as $\chi_{3}$ such that
$$
\chi_{3}(n)=\sum_{d \mid n} \chi_{1}(d) \chi_{2}\left(\frac{n}{d}\right)
$$
then it is well-known that multiplicative functions $(\chi(m) \chi(n)=\chi(m n)$ if $\operatorname{gcd}(m, n)$, so e.g. $\phi(n)$, the Euler totient function) convolve to a multiplicative function.
In particular, $f_{0}$ is a multiplicative function by definition (it is equivalent to only define it at prime powers then multiply), so the convolution of $f_{0}$ with itself is multiplicative. By definition of $a_{n}$, the convolution of $f_{0}$ with itself equals 1 at all prime powers. Thus by multiplicativity, it equals the constant function 1 everywhere.
Two final things to note: $f_{0}(1)=a_{0}=1>0$, and $f$ satisfying the conditions in the problem statement is indeed unique (proceed by induction on $n$ that $f(n)$ is determined uniquely and that the resulting algorithm for computing $f$ gives a well-defined function). Therefore $f_{0}$, satisfying those same conditions, must equal $f$.
At last, we have
$$
f\left(p^{2019}\right)=f_{0}\left(p^{2019}\right)=\frac{\binom{4038}{2019}}{4^{2019}}
$$
so
$$
f\left(2018^{2019}\right)=f\left(2^{2019}\right) f\left(1009^{2019}\right)=\frac{\binom{4038}{2019}^{2}}{4^{4038}}=\frac{\binom{4038}{2019}^{2}}{2^{8076}} .
$$