Functional Equation on Natural Numbers

Adding 1 to both sides of the functional equation: $f(xy) + 1 = f(x)f(y) + f(x) + f(y) + 1$ $f(xy) + 1 = (f(x) + 1)(f(y) + 1)$

Let $g(n) = f(n) + 1$. Then $g(xy) = g(x)g(y)$. For any prime $p$, $f(p) = 1 \implies g(p) = f(p) + 1 = 2$.

Prime factorization of $160000$: $160000 = 16 \times 10^4 = 2^4 \times (2 \times 5)^4 = 2^8 \times 5^4$

Using the multiplicative property of $g$: $g(160000) = g(2^8 \times 5^4) = (g(2))^8 \times (g(5))^4 = 2^8 \times 2^4 = 2^{12} = 4096$

Since $f(n) = g(n) - 1$: $f(160000) = 4096 - 1 = 4095$.