27

A number which is equal to the sum of the squares of its prime factors with multiplicity:

  • $16=2^2+2^2+2^2+2^2$
  • $27=3^2+3^2+3^2$

Are these the only two such numbers to exist?

There has to be an easy proof for this, but it seems to elude me.

Thanks

MJD
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barak manos
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  • So the general case would be $\prod p_k^{e_k}=\sum e_kp_k^2$? – Hagen von Eitzen Jan 10 '15 at 14:01
  • @HagenvonEitzen: Yes. – barak manos Jan 10 '15 at 14:04
  • For $e_k>1$ the left side grows exponentially, whereas the right hand side grows polynomially, so the solutions for $p_k$ are limited to small primes. – Lucian Jan 10 '15 at 15:39
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    There are no other solutions below $10^7$. – Lucian Jan 10 '15 at 15:48
  • @Lucian: Thanks. Can't currently really think of any other way to look into this besides exhaustively search. Perhaps the proof is not so easy after all, considering that nobody has given any suggestions up until now... – barak manos Jan 10 '15 at 15:54
  • You can start by proving that $16$ and $27$ are the only prime powers with this property, namely $p^k=kp^2\iff k=p^{k-2}$. Again, one side grows exponentially, whereas the other grows linearly, with regard to k $($obviously, $p\ge2)$, so its only possible solutions can be small values of k. – Lucian Jan 10 '15 at 16:01
  • @Lucian: And are you also implying that a prime power is the only possible type of solution? – barak manos Jan 10 '15 at 16:02
  • No. Merely that you should take it slowly, and start from the ground up. Then you'll move on to numbers that are the product of two prime powers. – Lucian Jan 10 '15 at 16:10
  • @Lucian: But I am right in the (now realized) fact that there's no trivial way to show it... right? I have actually posted this as a first step in trying to analyze a more complex question of mine, but now it seems that this one by itself is not that simple. Thanks. – barak manos Jan 10 '15 at 16:16
  • I am not a huge fan of “rigorous” proofs. As I already said twice before, one side grows exponentially, whereas the other one doesn't, so, if any solutions are to exist, they should be small. I believe the computer search took care of that. I'll leave the details to the specialists. – Lucian Jan 10 '15 at 16:33
  • It is not clear what is the problem? Rewrite the sum in the equation. Specify the look of this equation. Then we will solve it and to know when decisions can be. You only need to specify the type of the equation. Do not write the numbers is confusing. – individ Jan 10 '15 at 18:03
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    @individ: I am finding it very difficult to understand what you are trying to say. For example, what do you mean by "Then we will solve it and to know when decisions can be"??? I wrote the explicit numbers because they are the only ones for which the definition at the top of the question ("a number which is equal to the sum of the squares of its prime factors with multiplicity") obviously holds. – barak manos Jan 10 '15 at 18:18
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    @individ: In addition to that, I think that the numbers actually make it easier to understand (rather than confusing as you suggest), and I believe that most users here will agree with me on that. If anything, then the mathematical notation itself would make it less intuitive for most people. I could have phrased in a pure mathematical notation (as Hagen von Eitzen did at the first comment here), but I did not see any advantage in doing it this way. If a question can be written in a simple concise manner while keeping it mathematically accurate, then there is no reason why it shouldn't be. – barak manos Jan 10 '15 at 18:19
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    @Lucian: You are talking as if you had proved this thing. – TonyK Jan 11 '15 at 16:39
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    @individ: Your comment is incomprehensible to me! Have you read the question all the way through? It's not very long. – TonyK Jan 11 '15 at 16:40
  • @Engineer Toast: You've deleted your question, so I'm "leaving you a message" here instead. This question on Math-Overflow, although in a slightly different version than your question (applying square root on the result when it is a perfect square), might also be of interest to you. – barak manos Mar 13 '15 at 21:47
  • @TonyK 3 such numbers were found (see my answer) and Lucian was wrong after all... – ZaMoC Dec 18 '20 at 22:39

3 Answers3

6

Giorgos Kalogeropoulos has found 3 such numbers, each having more than 100 digits.
You can find these numbers if you follow the links in the comments of OEIS A339062 & A338093

or here https://www.primepuzzles.net/puzzles/puzz_1019.htm

So, such numbers exist! It is an open question if there are infinitely many of them...

ZaMoC
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2

Here is a suggestion:

For a start one could investigate under which assumptions about the sizes of $n$ and real variables $x_k\geq2$ $\>(1\leq k\leq n)$ an equality $$\prod_{k=1}^n x_k=\sum_{k=1}^n x_k^2$$ is at all possible.

Christian Blatter
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    Well, in order to "fit" into the question at hand, $x_k$ should be prime for every $1\leq{k}\leq{n}$ to begin with, right? Also, after thinking a little about @Lucian's comment, I have realized that it is not possible when $x_k$ is the same prime for all $1\leq{k}\leq{n}$ (i.e., the left side of your equation is a prime-power), except for the two cases already mentioned ($16$ and $27$). – barak manos Jan 11 '15 at 13:43
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    Though I admit that I have not dedicated any thought as to the value of $n$ (the total number of primes in the prime-factorization of the input number), for which this may be possible. It works with $n=4$ in the case of $16$ and $n=3$ in the case of $27$, but I don't see any correlation between these two cases. – barak manos Jan 11 '15 at 13:47
2

Maple confirms that $A$ and $B$ are prime, and that $N$ is a solution to the question. $N$ has 179 digits.
$$N=45AB\\ A=(C^{107}+D^{107})/(C+D)\\ B=(C^{109}+D^{109})/(C+D)\\ C=(\sqrt{47}+\sqrt{43})/2\\ D=(\sqrt{47}-\sqrt{43})/2$$ The general solution to $3^2+3^2+5^2+A^2+B^2=3×3×5×A×B$ is a sequence with recursion $$a_{n+1}=45a_n-a_{n-1}$$ The basic equation is a quadratic in any of the prime factors. The roots of the quadratic add up to the product of the rest of the prime factors. That is an easy way to generate solutions to $\sum x_i^2=\prod x_i$, although the $x_i$ might not be prime.
The starting point for this solution was $1,1,3,3,4$. Replace $4$ by $1×1×3×3-4=5$, then alternately replace the first factor $A$ by $B×3×3×5-A$ and the second factor $B$ by $A×3×3×5-B$.
Another starting point might be $1,1,1,2,2,2,3$ but I haven't found a prime solution from that.

Empy2
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