Main Question

What is wrong with this proof that there are no odd perfect numbers?

The "Proof"

Euler proved that an odd perfect number $N$, if any exists, must take the form $N = q^k n^2$ where $q$ is the Euler prime satisfying $\gcd(q,n)=1$ and $q \equiv k \equiv 1 \pmod 4$.

Denote the sum of divisors of $x \in \mathbb{N}$ by $\sigma(x)$, and the deficiency by $D(x)=2x-\sigma(x)$. Finally, denote the abundancy index by $I(x)=\sigma(x)/x$.

Consider the quantity $D(q^k)/\sigma(q^k)$: $$\dfrac{D(q^k)}{\sigma(q^k)}=\dfrac{D(q^k)}{q^k}\cdot\dfrac{1}{I(q^k)}=\dfrac{1}{I(q^k)}\cdot\bigg(2-I(q^k)\bigg)=I(n^2)-1.$$ Multiplying both sides by $n^2$, we get $$\dfrac{D(q^k)\cdot{n^2}}{\sigma(q^k)}=\sigma(n^2)-n^2 \in \mathbb{N}.$$ It is easy to show the following claims:

Claim 1 $\sigma(q^k) \nmid D(q^k)$

Proof Suppose to the contrary that $\sigma(q^k) \mid D(q^k)=2q^k - \sigma(q^k)$. This implies that $\sigma(q^k) \leq 2q^k - \sigma(q^k)$, which contradicts $q^k + 1 \leq \sigma(q^k)$. QED

Claim 2 $\sigma(q^k) \nmid n^2$

Proof Suppose to the contrary that $\sigma(q^k) \mid n^2$. Then $$\dfrac{n^2}{\sigma(q^k)}=\dfrac{\sigma(n^2)}{2q^k}$$ is an integer, contradicting the fact that $\sigma(n^2)$ is odd. QED

Hence, $\sigma(q^k) \nmid \bigg(D(q^k)\cdot{n^2}\bigg)$, contradicting $$\dfrac{D(q^k)\cdot{n^2}}{\sigma(q^k)}=\sigma(n^2)-n^2 \in \mathbb{N}.$$

  • 2
    $\begingroup$ See also here. $\endgroup$ – Dietrich Burde Mar 5 '17 at 19:25
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    $\begingroup$ Without studying the details, knowing that $a \, \nmid\, b $ and $a \, \nmid \, c $ does not in general imply that $a \, \nmid \, bc $. $\endgroup$ – lulu Mar 5 '17 at 19:55

$$D(q^k)\cdot{n^2} = \bigg(2q^k - \sigma(q^k)\bigg)\cdot{n^2} = 2{q^k}{n^2} - \sigma(q^k){n^2}$$ $$= \sigma(q^k)\sigma(n^2) - \sigma(q^k){n^2} = \sigma(q^k)\cdot\bigg(\sigma(n^2) - n^2\bigg)$$ which is divisible by $\sigma(q^k)$.

Therefore, there is no contradiction.


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