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Some New Results on Odd Perfect Numbers - Part IV

Continuing from this earlier blog post : (In particular, if $m$ is almost perfect, then $m$ must be a square (since $m$ is odd). Consequently, if the equation $\sigma(p^k)=2m$ holds (which is true if and only if $G=H=I$), then $\sigma(p^k)/2$ is also a square. This implies that $k=1$, since $$p^k + 1 \leq \sigma(p^k)=2m<2p$$ holds under the assumption $m<p$.) Summarizing our results so far, we have the chain of implications: $$m < p \Rightarrow \Bigg((m < p^k) \land (k = 1) \land (D(m) = 1)\Bigg).$$  But this blog post proves that the condition $D(m) = 1$ is equivalent to $m < p$. Since $\sigma(m^2)/p^k$ is also a square if $\sigma(p^k)/2$ is a square, then under the assumption that $m < p$, we obtain $\sigma(m^2)/p^k = \sigma(p^k)/2 = m$ is a square. (Note that the difference $m^2 - p^k$ is also a square.) Additionally, we obtain $$\sigma(p^k) = p^k + 1$$ since $k=1$ follows from the assumption $m < p$.  Since $m^2 - p^k$ is a square if and only if $$m^2 - p^k
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The journey of a thousand miles begins with a single step.

(Preamble: This blog post is lifted verbatim from https://arnienumbers101.blogspot.com/2023/05/a-journey-of-thousand-miles-begins-with.html .) https://rasputinnumbers.blogspot.com/2023/05/some-new-results-on-odd-perfect-numbers.html https://arnienumbers.blogspot.com/2023/05/dris-conjecture-holds-if-and-only-if.html In what follows, we will denote the classical sum of divisors of the positive integer $x$ by $$\sigma(x)=\sigma_1(x).$$ We will also denote the abundancy index  of $x$ by $I(x)=\sigma(x)/x$. We start with a minor technical lemma: Lemma 1: If $N = p^k m^2$ is an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, then $I(m^2) < \zeta(2)$ if and only if $p = 5$ and $k \neq 1$. Proof: Suppose that $I(m^2) < \zeta(2)$.  Since $2(p - 1)/p < I(m^2)$, then it follows that $$\frac{2(p - 1)}{p} < \zeta(2).$$ Hence, we derive $$p < \frac{12}{12 - {\pi}^2} < 6,$$ from which we infer that $p = 5$ (since $p$ is a
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Towards A Proof That There Are No Odd Perfect Numbers

#OPNResearch     #May2023 ( 1 )     https://math.stackexchange.com/questions/3779447 ( 2 )     https://math.stackexchange.com/questions/4690948 ( 3 )     https://math.stackexchange.com/questions/4692359 ( 4 )     https://mathoverflow.net/questions/406701 ( 5 )     https://mathoverflow.net/questions/404849 ( 6 )     https://math.stackexchange.com/questions/3061040 ( 7 )     https://arxiv.org/abs/2007.14305 ( 8 )     https://math.stackexchange.com/questions/3830794 ( 9 )     https://mathoverflow.net/questions/310919 ( 10 )   https://mathoverflow.net/questions/227076 ( 11 )   https://mathoverflow.net/a/447149/10365 For more information, please check this blog post .
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Some New Results on Odd Perfect Numbers - Part II

Let $N = p^k m^2$ be a hypothetical odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p, m) = 1$.  Note that, since $\gcd(p, m) = 1$ and $p$ is (the special) prime, then $p^k \neq m$.  (This also follows from the fact that prime powers are deficient, contradicting $N$ is perfect.)  By trichotomy, either $p^k < m \oplus m < p^k$ is true, where $\oplus$ denotes exclusive-OR.  (That is, $A \oplus B$ is true if and only if exactly one of $A$ or $B$ holds.) In what follows, we will denote the   classical sum of divisors  of the positive integer   z  by $\sigma(z)=\sigma_1(z)$ and the   abundancy index  of $z$ by $I(z)=\sigma(z)/z$. We will also denote the   deficiency  of $z$ by $D(z)=2z-\sigma(z)$, and the   aliquot sum   of $z$ by $s(z)=\sigma(z)-z$. I learned five (5) very important math lessons two days ago: (1)   From " The Abundancy Index of Divisors of Odd Perfect Numbers " (JIS, 09/2012) , we have the estimate $p^k < m
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