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.
If you have a product of integers $abc\cdots z \equiv 2 (\bmod{4})$ then you know the product is even. If all of $a,b,c,\ldots$ were odd then the product would be odd. So at least one of the integers is even. If more than one of the integers is even, then the product would be divisible by $4$ and so would be $\equiv 0 (\bmod{4})$ rather than $2$. So only one of the integers can be even. Moreover, if that integer were divisible by $4$, then the product, again, would be congruent to $0$, rather than $2$, so that integer must be divisible by $2$, but not $4$. Hence it is congruent to $2$ modulo $4.$
The answer to your second bold statement is similar. Of one factor is divisible by $4$, it makes the whole product divisible by $4$, so the whole product would be congruent to $0$, rather than $2$ modulo $4$.