Peter wrote: Show that $ \sigma (n) - d(m)$ is even for all positive integers $ m$ and $ n$ where $ m$ is the largest odd divisor of $ n$. $ n = 2^{k}m$ $ \sigma (2^km) = \sigma (2^k)\sigma (m)\equiv \sigma (m) (\mod 2)$ Let $ m = \prod_{i = 1}^{r} p_i^{a_i}$ where $ p\equiv 1 (\mod 2)$ We has $ \sigma (m) = \prod _{i = 1}^r\frac {p_i^{a_i + 1} - 1}{p - 1}\equiv (a_1 + 1)...(a_r + 1) = \tau(m)$ So $ \sigma(n)\equiv \tau(m)(\mod 2)$ This problem was be prove.

So the problem boils down to the trivial $1+t+t^2+\cdots +t^s \equiv s+1 \pmod{2}$ for odd $t$. Then $\sigma(n) =\sigma(2^km) = \sigma(2^k)\sigma(m) \equiv \sigma(m) \equiv \tau(m) \pmod{2}$, since $\sigma(m) = \prod_{p \textrm{ prime, } p\mid m} \sum_{i=0}^{v_p(m)} p^{i} \equiv $ $\prod_{p \textrm{ prime, } p\mid m} (v_p(m) + 1) =$ $ \tau(m) \pmod{2}$.