A function is defined on the set of positive integers such that if $$𝑛$$ is an odd integer, $$𝑓(𝑛) = 𝑛 − 1$$ and if $$𝑛$$ is an even integer, $$𝑓(𝑛) = n^{2} - 1$$. Determine the sum of all possible values of 𝑛 such that $$𝑓(𝑓(𝑛)) = 99$$.

First write the definition of the function.

If $$n$$ is odd, $$f(n)=n-1$$ (which is even).
If $$n$$ is even, $$f(n)=n^{2}-1$$ (which is odd).

We want all positive integers $$n$$ for which $$f(f(n))=99$$.

$$n$$ is odd.
Then $$f(n)=n-1$$ (even).
Apply $$f$$ again: $$f(f(n))=(n-1)^{2}-1$$ because the input is now even.
Set equal to $$99$$:

$$(n-1)^{2}-1 = 99 \;\;\Longrightarrow\;\; (n-1)^{2}=100 \;\;\Longrightarrow\;\; n-1=\pm10.$$

Since $$n$$ must be positive, $$n-1=10$$ gives $$n=11$$, while $$n-1=-10$$ gives the negative integer $$n=-9$$, which is rejected. Hence the only solution in this case is $$n=11$$.

$$n$$ is even.
Then $$f(n)=n^{2}-1$$ (odd).
Apply $$f$$ again: $$f(f(n))=(n^{2}-1)-1 = n^{2}-2.$$ Set equal to $$99$$:

$$n^{2}-2 = 99 \;\;\Longrightarrow\;\; n^{2}=101.$$

But $$101$$ is not a perfect square, so there is no even integer $$n$$ satisfying the equation.

Combining both cases, the only valid positive integer is $$n=11$$.

Required sum of all possible values of $$n$$ = $$11$$.

Answer: 11