# If $N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}}$ is an odd perfect number, then ${p_i}^{\alpha_i} < \sqrt{N}$.

In this paper, Peter Acquaah asserts that an important difference between odd perfect and even perfect numbers is that:

(A) The greatest component of an odd perfect number $N$ is less than $\sqrt{N}$.

(B) The greatest component of an even perfect number $M$ is greater than $\sqrt{M}$.

ATTEMPT TO PROVE STATEMENT (B)

Let $M = {2^{p-1}}(2^p - 1)$ be an even perfect number. Since $2^{p-1} < 2^p - 1$ and $2^p - 1$ is prime, the largest component of $M$ is the Mersenne prime $2^p - 1$.

We want to show that $$2^p - 1 > \sqrt{M} = 2^{\frac{p-1}{2}}\sqrt{2^p - 1}.$$ Assume to the contrary that $$2^p - 1 \leq 2^{\frac{p-1}{2}}\sqrt{2^p - 1}.$$ This implies that $$\sqrt{2^p - 1} \leq 2^{\frac{p-1}{2}}$$ which means $$2^p - 1 \leq 2^{p-1}.$$ This contradicts $2^{p-1} < 2^p - 1$.

Now here is my question:

How do you prove Statement (A)?

I only know that an odd perfect number (if it exists) must take the form $N = q^k n^2$ where $\gcd(q,n)=1$ and $q$ is prime (called the Euler prime) satisfying $q \equiv k \equiv 1 \pmod 4$. I also know that $$q^k < \frac{2}{3}n^2.$$ (See this paper for a proof.)

Any hints will be appreciated.

In particular, if $$N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}}$$ is the canonical factorization of an odd perfect number $N$, then one of the results in my thesis states that $$\sigma({p_i}^{\alpha_i}) \leq \frac{2}{3}\cdot\frac{N}{{p_i}^{\alpha_i}}$$ for all $i$, $1 \leq i \leq \omega(N)$. (This has since been continually improved by several authors.)
In particular, $${p_i}^{\alpha_i} < \sigma({p_i}^{\alpha_i}) \leq \frac{2}{3}\cdot\frac{N}{{p_i}^{\alpha_i}} < \frac{N}{{p_i}^{\alpha_i}}.$$ This implies that ${p_i}^{\alpha_i} < \sqrt{N}$ for all $i$.