(**Note:** The contents of this blog post were taken verbatim from this Mathematics Stack Exchange question. In this regard, kindly see the rebuttals from Professor Dujella in this chat room.)

Let $N$ be an odd perfect number given in the so-called *Eulerian form*

$$N = q^k n^2$$

where $q$ is the *special prime* satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

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In what follows, let us keep in mind the following lemma: If $q^k n^2$ is an odd perfect number given in Eulerian form, then $k = 1$ holds if and only if

LEMMA:

$$s(n^2) = \left(\frac{q-1}{2}\right)\cdot{D(n^2)},$$

where $s(x)=\sigma(x)-x$ is the

*aliquot sum*of the positive integer $x$, $D(x)=2x-\sigma(x)$ is the

*deficiency*of $x$, and $\sigma(x)=\sigma_1(x)$ is the

*classical sum of divisors*of $x$.

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Now, we start with

$$\gcd(n^2,\sigma(n^2)) = \left(q^k t - 2(q - 1)\right)n^2 + \left((-\sigma(q^k)/2)\cdot{t} + q\right)\sigma(n^2),$$

like Tony Kuria Kimani did in this recent ResearchGate preprint.

But the equation

$$\gcd(n^2,\sigma(n^2)) = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2} = \frac{s(n^2)}{D(q^k)/2} = \frac{D(n^2)}{s(q^k)}$$

holds. Consequently, we obtain the simultaneous equations

$$s(n^2) = (D(q^k)/2)\cdot\gcd(n^2,\sigma(n^2))$$

$$= \left(q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)\right)n^2 + \left((-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)\right)\sigma(n^2)$$

$$D(n^2) = {s(q^k)}\cdot\gcd(n^2,\sigma(n^2)) = \left(q^k s(q^k) t - 2(q - 1)s(q^k)\right)n^2 + \left((-\sigma(q^k)/2)s(q^k)\cdot{t} + qs(q^k)\right)\sigma(n^2).$$

We now test whether the equation

$$s(n^2) = \left(\frac{q-1}{2}\right)\cdot{D(n^2)}$$

holds; that is, whether the equation

$$\left(q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)\right)n^2 + \left((-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)\right)\sigma(n^2)$$

$$= \left(\frac{q-1}{2}\right)\cdot\Bigg(\left(q^k s(q^k) t - 2(q - 1)s(q^k)\right)n^2 + \left((-\sigma(q^k)/2)s(q^k)\cdot{t} + qs(q^k)\right)\sigma(n^2)\Bigg) \tag{1}$$

is true.

For simpler algebra, let

$$u_1 = q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)$$

$$u_2 = \left(\frac{q-1}{2}\right)\cdot\left(q^k s(q^k) t - 2(q - 1)s(q^k)\right)$$

$$v_1 = (-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)$$

and

$$v_2 = \left(\frac{q-1}{2}\right)\cdot\left((-\sigma(q^k)/2)s(q^k)\cdot{t} + qs(q^k)\right).$$

Then Equation $(1)$ becomes

$$\left(u_1 - u_2\right)\cdot{n^2} = \left(v_2 - v_1\right)\cdot{\sigma(n^2)}. \tag{2}$$

We now attempt to get simplified expressions for $u_1 - u_2$ and $v_2 - v_1$ using WolframAlpha. We obtain the following:

$$u_1 - u_2 = -\left(\frac{q^k - q}{2(q - 1)}\right)\cdot\left(tq^k - 2q + 2\right)$$

$$v_2 - v_1 = -\left(\frac{q^k - q}{2(q - 1)}\right)\cdot\left(\frac{q(tq^k - 2q + 2) - t}{2(q - 1)}\right). \tag{3}$$

Dividing both sides of Equation $(2)$ by $d = \gcd(n^2,\sigma(n^2))$, we obtain

$$\left(u_1 - u_2\right)\cdot\Bigg({\frac{n^2}{d}}\Bigg) = \left(v_2 - v_1\right)\cdot\Bigg(\frac{\sigma(n^2)}{d}\Bigg).$$

Recall that

$$d = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2}.$$

Consequently, we have

$$\left(u_1 - u_2\right)\cdot\Bigg(\frac{\sigma(q^k)}{2}\Bigg) = \left(v_2 - v_1\right)\cdot{q^k}. \tag{4}$$

Since $\gcd(q^k,\sigma(q^k)/2)=1$, then there exists an integer $G$ such that

$$v_2 - v_1 = G\cdot\left(\sigma(q^k)/2\right).$$

Substituting into Equation $(4)$, we get

$$\left(u_1 - u_2\right)\cdot\Bigg(\frac{\sigma(q^k)}{2}\Bigg) = G\cdot\left(\sigma(q^k)/2\right)\cdot{q^k},$$

from which we finally obtain

$$u_1 - u_2 = G\cdot{q^k}.$$

We therefore finally get

$$\gcd(u_1 - u_2, v_2 - v_1) = \gcd\Bigg(G\cdot{q^k}, G\cdot\left(\sigma(q^k)/2\right)\Bigg) = G\cdot\gcd(q^k, \sigma(q^k)/2) = G. \tag{5}$$

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To recap, we have obtained

$$G = \gcd(u_1 - u_2, v_2 - v_1).$$

However, from the equations in $(3)$, we also get

$$\gcd(u_1 - u_2, v_2 - v_1) = -\left(\frac{q^k - q}{2(q - 1)}\right).$$

Hence, we finally have

$$G = \gcd(u_1 - u_2, v_2 - v_1) = -\left(\frac{q^k - q}{2(q - 1)}\right).$$

Since GCDs are always nonnegative, then $G \geq 0$, which means that we have

$$q^k - q \leq 0.$$

This implies that $k \leq 1$. Since $k \geq 1$ ought to hold (because $k$ is a positive integer satisfying $k \equiv 1 \pmod 4$), then we now know that $k=1$.

Quite apparently, this implies that $G = 0$, whereupon we obtain $u_1 = u_2$ and $v_1 = v_2$.

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Here is my question:

Can I define GCDs to be * always positive* (so that $G > 0$ in the previous section) and thereby get a proof for $k \neq 1$?