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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?

Community
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TheSimpliFire
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  • 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? – Reiner Martin Nov 23 '17 at 21:11
  • @ReinerMartin possibly, since $2+8=10$ – TheSimpliFire Nov 23 '17 at 21:12
  • 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 $9k+1$ – Peter Nov 23 '17 at 21:35
  • The formulation of $a)$ is unlucky. Do you mean that every perfect square greatert than $4$ has such a representation ? – Peter Nov 23 '17 at 21:37
  • $b)$ is false because $10$ cannot be written in such a way. – Peter Nov 23 '17 at 21:55
  • But $b)$ seems to be true for $n\ge 11$, which would imply that $a)$ is true (If you mean that every perfect square greater than $4$ has such a representation). For $11\le n\le 1000$ , $b)$ is true. – Peter Nov 23 '17 at 22:16
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    $a)$ needs adjusting slightly. Since $qr\ge4$, and $qr<n$ is required, it doesn't work for $n=3,4$ – nickgard Nov 23 '17 at 22:44
  • @nickgard I have changed the statement in a). I meant both $q$ and $r$ must be less than $n$, not their product. – TheSimpliFire Nov 24 '17 at 07:38
  • @Peter edited to $n>10$. How would you go about proving b)? – TheSimpliFire Nov 24 '17 at 07:46
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    To prove such things is usually extremely difficult, Maybe partial results concerning the Goldbach-conjecture are helpful. – Peter Nov 25 '17 at 12:00
  • How is conjecture (d) true when $n=32$ (or for any larger even number up to $98$)? – David K Nov 25 '17 at 21:40
  • @simplyfire, supposedly they are your own discovered conjectures, do you offer the bounty for anyone finding the remaining proof ? or do you have a strategy to partition it ? – Abr001am Nov 26 '17 at 01:13
  • @Idle The bounty is offered to anyone who posts a complete proof to at least one of the above conjectures except c). I do not know how to partition a bounty but I guess upvoting and accepting (25 rep) will do the job. – TheSimpliFire Nov 26 '17 at 07:19
  • @DavidK You are right. I shall delete it. However, I'm interested in why it works for $n<30$. – TheSimpliFire Nov 26 '17 at 07:26
  • You could put (d) back, just change 100 to 30. – David K Nov 26 '17 at 14:57
  • I don't think it should be on the list since it only works for a few integers. I might make a separate post about it though. – TheSimpliFire Nov 26 '17 at 15:05
  • TheSimpliFire and @peter and all for the 2nd conjecture, i'm not saying it's a lemoine's but whoever solves it can solve lemoine's conjecture immediately . – Abr001am Nov 26 '17 at 15:48
  • @Idle thanks I did not know about Lemoine's Conjecture. Part b) of the list is probably a stronger version of it as it takes into account the even integers as well. – TheSimpliFire Nov 26 '17 at 16:17
  • @DavidK I have posted it here: https://math.stackexchange.com/questions/2538173/on-the-division-of-integers-of-the-form-nnn-frac-n2-frac-n2-frac-n2-by-pi – TheSimpliFire Nov 26 '17 at 16:29
  • I want to give it an upvote for a clever title :D – AlvinL Dec 02 '17 at 11:13

2 Answers2

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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.

Reiner Martin
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  • Yes, I think this needs to be summed again to give $10$. How would you write the statement? – TheSimpliFire Nov 23 '17 at 21:14
  • @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 $1$. – Peter Nov 23 '17 at 21:51
  • Thanks for the link! – TheSimpliFire Nov 24 '17 at 07:48
  • It's worth emphasising that this only shows (c) for even perfect numbers - the linked article says nothing about odd perfect numbers. – B. Mehta Nov 28 '17 at 03:20
  • Good point. As a further addendum, note that it is not known if any odd perfect numbers exist, and there are none below $10^{1500}.$ – Reiner Martin Nov 28 '17 at 09:33
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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.

actinidia
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