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

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

• See also here. – Dietrich Burde Mar 5 '17 at 19:25
• Without studying the details, knowing that $a \, \nmid\, b$ and $a \, \nmid \, c$ does not in general imply that $a \, \nmid \, bc$. – 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 $\sigma(q^k)$.