Relating a sum of an even number of odd, positive integers to a power of two

Question

I was recently walking myself through a proof by contradiction that was premised on producing a result of the form $\frac{\text{odd}}{\text{even}}\implies\; \text{even} $, which, of course, is contradictory. I will spare the details here, but my question now is on how to relate the sum of an even number of positive, odd integers to a power of two. More precisely, suppose $\lambda$ is odd and not a perfect square with $$\lambda=\prod_{i=1}^kp_1^{a_i}$$ for all $p_i$ distinct primes and all $a_i\geq 1$. Let $d_i$ denote a divisor of $\lambda$ (implying all $d_i$ are odd). If we then take $$\sum_{i=1}^m d_i$$ (the sum of all divisors of $\lambda$) then this sum will be even because $\lambda$ is not a perfect square (implying an even number of terms) and because each divisor is odd. My question is if we denote $\sum_{i=1}^m d_i=M2^v$, where $M$ is odd, what can we say about $v$, the power of two? In other words, is it possible to assign a maximum value for $v$? A minimum? I am not sure what insight is necessary here, so I thought I'd ask.

Answer 1

If you want to make determinations about $\nu$, it might be useful to have $\nu$ be a function of the prime factors $p_k$ of $\lambda$. To illustrate with two examples, let $\lambda_0 = 17$; then $\nu_0 = 1$. And if $\lambda_1=19$, then $\nu_1 = 2$. Even with single prime factors, you can have different divisor sums.

The divisor summation function for a Mersenne Prime $M_n = 2^n - 1$ is $\sigma_1(M_n) = 2^n$, so you can have arbitrarily large $\nu$ for a single prime.