# On odd perfect numbers $q^k n^2$ and the deficient-perfect divisor $q^{frac{k-1}{2}} n^2$

(Note: This post is an offshoot of the following earlier question.)

Let $$sigma(x)$$ be the sum of divisors of the positive integer $$x$$. Denote the deficiency of $$x$$ by $$D(x)=2x-sigma(x)$$, and the sum of the aliquot divisors of $$x$$ by $$s(x)=sigma(x)-x$$.

If $$sigma(m)=2m$$ and $$m$$ is odd, then $$m$$ is called an odd perfect number. It is currently unknown whether there are any odd perfect numbers, although it is widely believed that there is none.

Euler proved that an odd perfect number $$m$$, if one exists, must have the so-called Eulerian form
$$m = q^k n^2$$
where $$q$$ is the special/Euler prime satisfying $$q equiv k equiv 1 pmod 4$$ and $$gcd(q,n)=1$$.

Since the divisor sum function $$sigma$$ is a multiplicative function and $$m = q^k n^2$$ is perfect, we obtain
$$2 q^k n^2 = 2m=sigma(m)=sigma(q^k n^2)=sigma(q^k)sigma(n^2)$$
so that we get
$$frac{sigma(n^2)}{q^k}=frac{n^2}{sigma(q^k)/2}=frac{2s(n^2)}{D(q^k)}=frac{D(n^2)}{s(q^k)}=gcd(n^2,sigma(n^2)).$$

From the hyperlinked MSE question, letting
$$x = q^{frac{k-1}{2}},$$
$$y = n^2,$$
and noting that
$$gcd(x,y)=gcd(q,n)=1$$
then we obtain
$$Dbigg(q^{frac{k-1}{2}}bigg)D(n^2) – Dbigg(q^{frac{k-1}{2}} n^2bigg) = 2sbigg(q^{frac{k-1}{2}}bigg)s(n^2).$$
But we know that
$$sigma(n^2)={n^2}cdotfrac{2q^k}{sigma(q^k)}$$
$$Dbigg(q^{frac{k-1}{2}}bigg)=2q^{frac{k-1}{2}} – sigmabigg(q^{frac{k-1}{2}}bigg)=2q^{frac{k-1}{2}} – frac{q^{frac{k+1}{2} – 1}}{q – 1}=frac{q^{frac{k+1}{2}} – 2q^{frac{k-1}{2}} + 1}{q-1}$$
$$D(n^2)=2n^2 – sigma(n^2)={2n^2}cdotbigg(frac{sigma(q^k)-q^k}{sigma(q^k)}bigg)={2n^2}cdotbigg(frac{q^k – 1}{(q – 1)sigma(q^k)}bigg)$$
$$Dbigg(q^{frac{k-1}{2}} n^2bigg)=frac{2 q^{frac{k-1}{2}} n^2}{q^{frac{k+1}{2}} + 1}$$
(The last equation is a result of Holdener and Rachfal.)
$$sbigg(q^{frac{k-1}{2}}bigg)=sigmabigg(q^{frac{k-1}{2}}bigg)-q^{frac{k-1}{2}}=sigmabigg(q^{frac{k-3}{2}}bigg)=frac{q^{frac{k-1}{2}} – 1}{q – 1}$$
$$s(n^2)=sigma(n^2)-n^2={n^2}cdotbigg(frac{2q^k – sigma(q^k)}{sigma(q^k)}bigg)={n^2}cdotbigg(frac{q^{k+1} – 2q^k + 1}{(q-1)sigma(q^k)}bigg)$$
Cancelling $$2n^2$$ from both sides of
$$Dbigg(q^{frac{k-1}{2}}bigg)D(n^2) – Dbigg(q^{frac{k-1}{2}} n^2bigg) = 2sbigg(q^{frac{k-1}{2}}bigg)s(n^2)$$
we obtain
$$bigg(frac{q^{frac{k+1}{2}} – 2q^{frac{k-1}{2}} + 1}{q-1}bigg)bigg(frac{q^k – 1}{(q – 1)sigma(q^k)}bigg)-frac{q^{frac{k-1}{2}}}{q^{frac{k+1}{2}} + 1}=bigg(frac{q^{frac{k-1}{2}} – 1}{q – 1}bigg)bigg(frac{q^{k+1} – 2q^k + 1}{(q-1)sigma(q^k)}bigg).$$

Now, assume that the Descartes-Frenicle-Sorli Conjecture that $$k=1$$ is true.

Then the first factor on the RHS of the last equation evaluates to zero, so that we have (on the LHS)
$$frac{1}{q+1}-frac{1}{q+1}=0.$$

However, when I performed an exact computation on both LHS and RHS of the equation (after cancelling $$2n^2$$ and multiplying both sides by $$sigma(q^k)$$), I am getting:

$$text{LHS} = frac{4q^{frac{3k+1}{2}}-2q^{frac{3k+3}{2}}-q^{k+1}-2q^{frac{3k-1}{2}}+q^{2k+1}+2q^k-2q^{2k} + q^k – 1}{bigg(q^{frac{k+1}{2}} + 1bigg)(q-1)^2}$$
$$text{RHS} = frac{-q^{k+1}+q^{frac{k-1}{2}}-2q^{frac{3k-1}{2}}+q^{frac{3k+1}{2}} + 2q^k – 1}{(q-1)^2}.$$

Assuming the veracity of the expressions for
$$text{LHS} = bigg(frac{q^{frac{k+1}{2}} – 2q^{frac{k-1}{2}} + 1}{q-1}bigg)bigg(frac{q^k – 1}{q – 1}bigg)-frac{q^{frac{k-1}{2}} sigma(q^k)}{q^{frac{k+1}{2}} + 1}$$
and
$$text{RHS} = bigg(frac{q^{frac{k-1}{2}} – 1}{q – 1}bigg)bigg(frac{q^{k+1} – 2q^k + 1}{q-1}bigg),$$
upon cancelling $$(q-1)^2$$ from both sides, we obtain
$$frac{4q^{frac{3k+1}{2}}-2q^{frac{3k+3}{2}}-q^{k+1}-2q^{frac{3k-1}{2}}+q^{2k+1}+2q^k-2q^{2k} + q^k – 1}{bigg(q^{frac{k+1}{2}} + 1bigg)}$$
$$=-q^{k+1}+q^{frac{k-1}{2}}-2q^{frac{3k-1}{2}}+q^{frac{3k+1}{2}} + 2q^k – 1.$$
Then, with a goal of simplifying the computations, again assuming that the Descartes-Frenicle-Sorli Conjecture that $$k=1$$ is true, we have
$$k=1 implies frac{4q^2 -2q^3 -q^2 -2q + q^3 + 2q -2q^2 + q – 1}{q + 1}=-q^2+1-2q+q^2+2q-1=0 implies -q^3 + q^2 + q – 1 = q^2 (1 – q) – (1 – q) = 0 = (q^2 – 1)(1 – q) = 0.$$
The last equation implies that
$$q = pm 1,$$
which contradicts $$q geq 5$$, since $$q$$ is a prime satisfying $$q equiv 1 pmod 4$$.

Here are my questions:

(1) Are the expressions
$$text{LHS} = bigg(frac{q^{frac{k+1}{2}} – 2q^{frac{k-1}{2}} + 1}{q-1}bigg)bigg(frac{q^k – 1}{q – 1}bigg)-frac{q^{frac{k-1}{2}} sigma(q^k)}{q^{frac{k+1}{2}} + 1}$$
$$= frac{4q^{frac{3k+1}{2}}-2q^{frac{3k+3}{2}}-q^{k+1}-2q^{frac{3k-1}{2}}+q^{2k+1}+2q^k-2q^{2k} + q^k – 1}{bigg(q^{frac{k+1}{2}} + 1bigg)(q-1)^2},$$
and
$$text{RHS} = bigg(frac{q^{frac{k-1}{2}} – 1}{q – 1}bigg)bigg(frac{q^{k+1} – 2q^k + 1}{q-1}bigg)$$
$$= frac{-q^{k+1}+q^{frac{k-1}{2}}-2q^{frac{3k-1}{2}}+q^{frac{3k+1}{2}} + 2q^k – 1}{(q-1)^2}$$
algebraically correct?

(2) If the answer to Question (1) is YES, how do we resolve the apparent contradiction?

Postscript: I used WolframAlpha to simplify some of the expressions in the interim.

Mathematics Asked by Jose Arnaldo Bebita Dris on November 21, 2021

Here is the WolframAlpha computational verification for the numerator of the $$text{LHS}$$ expression under consideration.

The computation simplifies the expression for $$text{LHS}$$ to $$text{LHS}=frac{-q^{k+1}+3q^{frac{3k+1}{2}}-q^{frac{k+1}{2}}+q^{frac{k-1}{2}}-2q^{frac{3k-1}{2}}-q^{frac{3k+3}{2}}+q^{2k+1}+3q^k-2q^{2k}-1}{bigg(q^{frac{k+1}{2}} + 1bigg)(q-1)^2}$$ which does not agree with the prior computation above.

Likewise, here is the WolframAlpha computational verification for the numerator of the $$text{RHS}$$ expression under consideration.

The computation simplifies the expression for $$text{RHS}$$ to $$text{RHS}=frac{-q^{k+1}+q^{frac{k-1}{2}}-2q^{frac{3k-1}{2}}+q^{frac{3k+1}{2}}+2q^k-1}{(q-1)^2}$$ which agrees with the prior computation above.

Equating $$text{LHS}=text{RHS}$$ and cancelling $$(q-1)^2$$ from both sides, we obtain $$frac{-q^{k+1}+3q^{frac{3k+1}{2}}-q^{frac{k+1}{2}}+q^{frac{k-1}{2}}-2q^{frac{3k-1}{2}}-q^{frac{3k+3}{2}}+q^{2k+1}+3q^k-2q^{2k}-1}{q^{frac{k+1}{2}} + 1}$$ $$=-q^{k+1}+q^{frac{k-1}{2}}-2q^{frac{3k-1}{2}}+q^{frac{3k+1}{2}}+2q^k-1.$$

Finally, here is the WolframAlpha computational verification for the product $$bigg(q^{frac{k+1}{2}} + 1bigg)bigg(-q^{k+1}+q^{frac{k-1}{2}}-2q^{frac{3k-1}{2}}+q^{frac{3k+1}{2}}+2q^k-1bigg).$$ The computation comes out to $$-q^{k+1}+2q^{frac{3k+1}{2}}-q^{frac{3k+3}{2}}+q^{frac{k-1}{2}}-q^{frac{k+1}{2}}-2q^{frac{3k-1}{2}}+q^{frac{3k+1}{2}}+q^k-2q^{2k}+q^{2k+1}+2q^k-1$$ $$=-q^{k+1}+3q^{frac{3k+1}{2}}-q^{frac{3k+3}{2}}+q^{frac{k-1}{2}}-q^{frac{k+1}{2}}-2q^{frac{3k-1}{2}}+3q^k-2q^{2k}+q^{2k+1}-1,$$ which agrees with the numerator of the $$text{LHS}$$.

Hence, there is actually NO CONTRADICTION, as the equation is in fact an IDENTITY.

Answered by Jose Arnaldo Bebita Dris on November 21, 2021

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