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Question: Show that all even perfect numbers end in 6 or 8.

This is what I have. All even perfect numbers are of the form $n=2^{p-1}(2^p -1)$ where $p$ is prime and so is $(2^p -1)$.

What I did was set $2^{p-1}(2^p -1)\equiv x\pmod {10}$ and proceeded to show that $x=6$ or $8$ were the only solutions.

Now, $2^{p-1}(2^p -1)\equiv x\pmod {10} \implies 2^{p-2}(2^p -1)\equiv \frac{x}{2}\pmod {5}$, furthermore there are only two solutions such that $0 \le \frac{x}{2} < 5$. So I plugged in the first two primes and only primes that satisfy. That is if $p=2$ then $\frac{x}{2}=3$ when $p=3$ then $\frac{x}{2}=4$. These yield $x=6$ and $x=8$ respectively. Furthermore All solutions are $x=6+10r$ or $x=8+10s$.

I would appreciate any comments and or alternate approaches to arrive at a good proof.

  • 0
    Why is it that if suffices to check $p=2$ and $p=3$? There are other values of $p$ that yield perfect numbers, how do you know they won't give you $x=1$ or $x=2$?2012-07-09
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    I was thinking I only need to consider the two $p$s that will give me the least residues.2012-07-09
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    *How* do you know that the smallest primes will give you the smallest residues? How do you know there is no large prime $p$ such that $2^{p-2}(2^p-1)\equiv 1\pmod{5}$? A larger number may have a smaller residue: $13$ has a smaller residue modulo $5$ than $4$.2012-07-09
  • 0
    I see what you are saying. Thanks. I am going to go back to work it out.2012-07-09

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