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The following three statements are my own conjectures, not a homework problem.

$a)$ For $n = 3, 4, 5,..$, every square integer $n^2$ can be expressed as the sum of a prime $p$ and two other primes $q$ and $r$ multiplied together with $q, r < n$.

$b)$: (Similar to $a)$) Every positive integer $n>10$ can be written as a sum of a prime $p$ and two other primes (not necessarily distinct) $q$ and $r$ multiplied together.

$c)\,\mathbf{[proven]}$: The digital root of every perfect number except $6$ is $1$.

Can you prove or disprove them? If this is difficult, are there any implications between $a),b)$ and Goldbach's conjecture?

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  • $\begingroup$ The conjecture (c) as written is false. $n=2^{p-1}(2^p-1)$ is a perfect number for 9941, but the sum of the digits of the sum of the digits is 28. Need to iterate until a single digit is reached? $\endgroup$ – Reiner Martin Nov 23 '17 at 21:11
  • $\begingroup$ @ReinerMartin possibly, since $2+8=10$ $\endgroup$ – TheSimpliFire Nov 23 '17 at 21:12
  • $\begingroup$ The digital root of a number is just the residue modulo $9$, unless the number is divisible by $9$, so to show $c)$, we only have to show that every perfect even number greater than $6$ is of the form $9k+1$ $\endgroup$ – Peter Nov 23 '17 at 21:35
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    $\begingroup$ $a)$ needs adjusting slightly. Since $qr\ge4$, and $qr<n$ is required, it doesn't work for $n=3,4$ $\endgroup$ – nickgard Nov 23 '17 at 22:44
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    $\begingroup$ To prove such things is usually extremely difficult, Maybe partial results concerning the Goldbach-conjecture are helpful. $\endgroup$ – Peter Nov 25 '17 at 12:00
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The conjecture (c) as written is false. $n=2^{p−1}(2^p−1)$ is a perfect number for $p=2203,$ but the sum of the digits of the sum of the digits of $n$ is 19.

However, if you keep iterating you arrive at 10 it seems. In other words, the digital root seem to be 1.

Actually, this is correct (all even perfect numbers other than 6 have digital root 1) and has been proven in http://apfstatic.s3.ap-south-1.amazonaws.com/s3fs-public/09-comac_digital_roots_of_perfect_numbers%20(1).pdf.

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  • $\begingroup$ Yes, I think this needs to be summed again to give $10$. How would you write the statement? $\endgroup$ – TheSimpliFire Nov 23 '17 at 21:14
  • $\begingroup$ @TheSimpliFire In fact it is not difficult to show that every perfect even number greater than $6$ is of the form $9k+1$ and hence if we iterate the sum of digits, we eventually reach the digitial root, which must be $1$. $\endgroup$ – Peter Nov 23 '17 at 21:51
  • $\begingroup$ Thanks for the link! $\endgroup$ – TheSimpliFire Nov 24 '17 at 7:48
  • $\begingroup$ It's worth emphasising that this only shows (c) for even perfect numbers - the linked article says nothing about odd perfect numbers. $\endgroup$ – B. Mehta Nov 28 '17 at 3:20
  • $\begingroup$ Good point. As a further addendum, note that it is not known if any odd perfect numbers exist, and there are none below $10^{1500}.$ $\endgroup$ – Reiner Martin Nov 28 '17 at 9:33
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(b) is an open problem.

If it weren't, then we would have resolution of both Lemoine's conjecture$^{[1]}$ and Marnell's prime and semiprime conjecture$^{[2]}$ (for which no exceptions below $1 \times 10^8$ exist).


[1] https://en.wikipedia.org/wiki/Lemoine%27s_conjecture

[2] Geoffrey R. Marnell, "Ten Prime Conjectures", Journal of Recreational Mathematics 33:3 (2004-2005), pp. 193--196.

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