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Two hours ago, I received a phone call from professor Halfbrain. The professor told me that he has detected an odd perfect number. The professor was very excited, since the existence of odd perfect numbers is an outstanding open question in number theory. Let me explain this question to you.

Definition: An integer $N$ is perfect, if the sum of all divisors of $N$ below $N$ exactly equals $N$.

For instance $N=6$ is perfect, as $1+2+3=6$. And also the number $N=28$ is perfect, as $1+2+4+7+14=28$.

Theorem: For a number $N$ with prime factorization $N=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ the sum $\sigma(N)$ of all divisors (including the number $N$ itself) is given by the expression $$ (1+p_1+p_1^2+\cdots+p_1^{e_1})\, (1+p_2+p_2^2+\cdots+p_2^{e_2})\, \cdots (1+p_k+p_k^2+\cdots+p_k^{e_k}). $$

The theorem implies that an integer $N$ is perfect, if and only if $\sigma(N)=2N$ holds true. With the help of the theorem, one easily checks and verifies that $6$ and $28$ are perfect:

  • $6=2\cdot3~~$ yields $~~\sigma(6)=(1+2)(1+3)=12=2\cdot6$
  • $28=2^2\cdot7~~$ yields $~~\sigma(28)=(1+2+4)(1+7)=56 =2\cdot 28$

To the current day, mathematicians have been hunting for odd perfect numbers. Although thousands of hours of research time have been invested into this problem, all these hours were of no avail. The big breakthrough of Professor Halfbrain on 8 November 2015 might be a turning point in the history of mathematics.

Professor Halfbrain's odd perfect number theorem:
For $N=3^2\cdot7^2\cdot11^2\cdot13^2\cdot22021 ~=~ 198,585,576,189$, we have
\begin{eqnarray} \sigma(N) &=& (1+3+3^2)(1+7+7^2)(1+11+11^2)(1+13+13^2)(1+22021) \\ &=& 397,171,152,378 ~~=~~ 2\cdot 198,585,576,189 ~~=~~ 2N. \end{eqnarray} Therefore $N=198,585,576,189$ constitutes an odd perfect number.

Question: Has the professor indeed managed to detect an odd perfect number, or is there a mistake hidden somewhere in the above paragraphs?

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The professor ...

... has made a mistake. $22021$ is not prime, it's $19^2 \cdot 61$.

\begin{align*} \small{ \sigma(N) } &= \small{ (1+3+3^2)(1+7+7^2)(1+11+11^2)(1+13+13^2)(1+19+19^2)(1+61) } \\ &= \small{ 426,027,470,778 \gt 2N } \end{align*}

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What's suspicious ...

... If you factor the individual products, $1 + 3 + 3^2 = 13$, $1 + 7 + 7^2 = 57 = 3 \cdot 19$, $1 + 11 + 11^2 = 133 = 7 \cdot 19$, $1 + 13 + 13^2 = 183 = 3 \cdot 61$, so there are a few factors not accounted for. It would seem that the two 19's and the 61 must be factors of 22021.

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