[Number Theory]
Problem: Let T be an integer. Given a nonzero integer $n$, let $f(n)$ denote the sum of all numbers of the form $i^{d}$, where $i=\sqrt{-1}$, and $d$ is a divisor (positive or negative) of $n$. Compute $f(2 T+1)$.

Solution: Let $n=2^{m} r$, where $r$ is odd. If $m=0$, then $n$ is odd, and for each $d$ that divides $n$, $i^{d}+i^{-d}=i^{d}+\frac{i^{d}}{\left(i^{2}\right)^{d}}=0$, hence $f(n)=0$ when $n$ is odd. If $m=1$, then for each $d$ that divides $n, i^{d}+i^{-d}$ equals 0 if $d$ is odd, and -2 if $d$ is even. Thus when $n$ is a multiple of 2 but not 4 , $f(n)=-2 P$, where $P$ is the number of positive odd divisors of $n$. Similarly, if $m=2$, then $f(n)=0$, and in general, $f(n)=2(m-2) P$ for $m \geq 1$. Because $T$ is an integer, $2 T+1$ is odd, hence the answer is $\mathbf{0}$. [Note: If $r=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdot \ldots \cdot p_{k}^{a_{k}}$, where the $p_{i}$ are distinct odd primes, it is well known that $P=\left(a_{1}+1\right)\left(a_{2}+1\right) \ldots\left(a_{k}+1\right)$.]

Final answer: 0