Problem: If $n$ is an odd perfect number then $n=p_1^{j_1}p_2^{2j_2}...p_r^{2j_r}$ where $p_i$ are distinct odd primes and $p_1\equiv j_1 \equiv 1 \pmod 4.$
I am trying to understand the proof due to Euler, given in Elementary Number Theory (Burton).
Let $n=p_1^{k_1}p_2^{k_2}...p_r^{k_r}.$ Since $n$ is perfect therefore $\sigma(n)=2n.$ Also since $n$ is odd $n\equiv 1\pmod 4$ or $n\equiv 3\pmod 4.$ In any case $2n\equiv 2\pmod 4\Rightarrow \sigma(n)\equiv2\pmod 4.$ Now $$\sigma(n)=\sigma(p_1^{k_1})\sigma(p_2^{k_2})...\sigma(p_r^{k_r})\equiv 2\pmod 4.$$ The implication is that one of the $\sigma(p_i^{k_i}),$ say $\sigma(p_1^{k_1})$ must be even but not divisible by $4$.
I don't understand why the claim stated in bold must be true.
Moving on,
For a given $p_i$, $p_i\equiv 1\pmod 4$ or $p_i\equiv -1\pmod 4.$ If $p_i \equiv -1\pmod 4$, then $\sigma(p_i^{k_i})\equiv 0,1\pmod 4$ depending on $k_i$ being odd or even respectively. Since $\sigma(p_1^{k_1})\equiv 2\pmod 4\Rightarrow p_1\equiv 1\pmod 4.$
Furthermore, $\sigma(p_i^{k_i})\equiv 0\pmod 4$ is not possible since this would imply that $4|\sigma(p_i^{k_i})$ which is not possible.
How can this be deduced?
The remainder of the proof is easy to understand. I am just facing trouble with these two parts. It would be nice if someone could elaborate more on these claims, maybe with an example.