# On prime factors of odd perfect numbers

Why is it so difficult to determine actual numerical values for prime factors of odd perfect numbers?

Recall that, if $$N$$ is an odd perfect number, then Euler proved that it takes the form $$N = q^k n^2$$ where $$q$$ is the special/Euler prime satisfying $$q \equiv k \equiv 1 \pmod 4$$ and $$\gcd(q, n) = 1$$. Also, it is not too computationally tedious to verify that $$3 \cdot 5 \cdot 7 = 105 \nmid N$$.

Lastly, note that it is currently unknown whether $$q = 5$$ holds.

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@GerryMyerson: Thank you for your comment. For example: "Assuming the Descartes-Frenicle-Sorli conjecture that $$k=1$$, then $$n^2$$ is deficient-perfect for which we only know of the single example $$n = 3 \cdot 7 \cdot {11} \cdot {13}$$. This gives rise to the Descartes spoof $$22021 \cdot n^2$$ which is not perfect since $$22021$$ is not prime." So essentially, my question is: What would happen if $$k \neq 1$$?

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@GerryMyerson: It is known (by work of Holdener and Rachfal) that $$q^{\frac{k-1}{2}} n^2$$ is deficient-perfect, if $$k \neq 1$$. But then, notice that the odd perfect number takes the form $$N = q^{\frac{k+1}{2}} \bigg(q^{\frac{k-1}{2}} n^2 \bigg)$$ and that $$\gcd\bigg(q^{\frac{k+1}{2}}, q^{\frac{k-1}{2}} n^2 \bigg) = q^{\frac{k-1}{2}} \neq 1.$$

• I'm having trouble imagining what kind of answer you could hope to get to this question. – Gerry Myerson May 26 at 9:07
• @GerryMyerson: Thank you for your comment. For example: "Assuming the Descartes-Frenicle-Sorli conjecture that $k=1$, then $n^2$ is deficient-perfect, for which we only know of the single example $n = 3 \cdot 7 \cdot {11} \cdot {13}$. This gives rise to the Descartes spoof $22021 \cdot n^2$ which is not perfect since $22021$ is not prime." So essentially, my question is: What would happen if $k \neq 1$? – Jose Arnaldo Bebita-Dris May 26 at 9:14
• Still not following. What does "What would happen" mean? What would happen to what if $k\ne1$? – Gerry Myerson May 27 at 2:06
• @GerryMyerson: It is known (by work of Holdener and Rachfal) that $$q^{\frac{k-1}{2}} n^2$$ is deficient-perfect, if $k \neq 1$. But then, notice that the odd perfect number takes the form $$N = q^{\frac{k+1}{2}} \bigg(q^{\frac{k-1}{2}} n^2 \bigg)$$ and that $$\gcd\bigg(q^{\frac{k+1}{2}}, q^{\frac{k-1}{2}} n^2 \bigg) = q^{\frac{k-1}{2}} \neq 1.$$ – Jose Arnaldo Bebita-Dris May 27 at 4:26

Probably, because in general they don't really seem to have much special in their factorization other than a very abstract form. ( like ex. Mersenne numbers of prime exponents do, which if you want to see more of, you might want to check out mersenne.org as GIMPS are heading towards fudiciary troubles apparently). Mersennes of prime exponent $$p$$ are of form $$2kp+1$$ and so are any divisors. More so, they have a lot of congruences to eliminate candidate factors quickly.
Odd perfect numbers have very little information built up, about what general factor looks like. That means we basically have to try them all. There are only 14680064 possibilities for distinct w-almost primes left, using primes under 100 given your above information about 105 not dividing. Only 8192 of these force anything congruent 3 mod 4 on $$n$$ at the moment ( okay directly without primes that may be $$q$$), using your own comments.
If I want to find guaranteed factors of say $$M_{8191}$$, I know the smallest exceeds $$M_{16}$$ right off the start. Number of candidates without sieving, only doubles every bit level. Assuming I try randomly, that equates your needed testing with testing this Mersenne number up to 20 bits in depth.