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Determine all one to one functions $f:\mathbb{N}^* \rightarrow \mathbb{N}^*$ (where $\mathbb{N}^*$ means all positive integers) having the following property:

For all $S$, where $S$ is a finite set of positive integers so that: $$ \sum_{s \in S} \frac{1}{s} \in \mathbb{N}^* $$ it implies: $$ \sum_{s \in S} \frac{1}{f(s)}\in \mathbb{N}^* $$

Of course, the identical function is a solution, but how about other solutions?

Update

I was able to prove (with help from a friend) that $f(n)=n, \forall n$ using induction and:

Egyptian fractions theorem. For every positive rational r and positive integer N, there exists a set $ \{ n_1, . . . , n_k\}$ of positive integers such that $n_i > N$ for every $i = 1, 2, . . . , k$ and $$r = \sum_{1\le i \le k}\frac {1}{n_i}$$

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    What do you mean by $N^*$? – Erick Wong Aug 19 '15 at 13:47
  • @Erick Wong I updated the question –  Aug 19 '15 at 13:52
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    Is $S$ necessarily a set of distinct positive integers? (that is, are you allowing things like $\frac 12 + \frac 12=1$). – lulu Aug 19 '15 at 13:52
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    @lulu A set, more or less by definition, has only distinct elements. If you want to allow more than one of any given element, then it's usually called a multiset. – Arthur Aug 19 '15 at 13:54
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    @lulu every set has distinct elements –  Aug 19 '15 at 13:54
  • @EugenCovaci Are there actually many sets $S$ that satisfy the condition? Could you provide some examples for such sets? I could only come up with ${2, 3, 6}$.. – Dominik Aug 30 '15 at 15:45
  • @Dominik Some clever examples are given by perfect numbers, like $6$, but there are many more. See also Find five positive integers whose reciprocals sum to $1$. – Bart Michels Aug 30 '15 at 15:58
  • @Dominik See Egyptian fractions. It's trivial to create infinitely many examples by splitting $\frac1n \mapsto \frac1{n+1} + \frac1{n(n+1)}$, but there are many many more than just this. – Erick Wong Aug 30 '15 at 16:04
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    Note that $f(1)=1$. The only non-trivial solution to $\frac1a+\frac1b+\frac1c\in\Bbb N$ is $(2,3,6)$ and permutations, so that gives us some information to start with... – Bart Michels Aug 30 '15 at 16:08
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    For four numbers the only possibilities are $(2,3,7,42),(2,3,8,24),(2,3,9,18),(2,4,5,20),(2,4,6,12)$. By the way, does anyone know where those solutions are listed for a small number of variables (say at most $10$ or so)? – Bart Michels Aug 30 '15 at 18:27
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    @barto Ah, so that shows $f(2) = 2$ (since it's the only number occurring in every $4$-tuple, from which follows $f(3) = 3, f(6) = 6$. This does lend weight to the hypothesis that the identity map is the only solution. This would be easy if one could prove something like "if $m < n$ then for some sufficiently large $k$ there are more $k$-tuples containing $m$ than there are containing $n$". – Erick Wong Aug 31 '15 at 01:44
  • The posibilities for lenght = 5 are (2, 3, 8, 40, 60) (2, 3, 8, 42, 56) (2, 3, 9, 27, 54) (2, 3, 9, 30, 45) (2, 3, 10, 20, 60) (2, 3, 10, 24, 40) (2, 3, 11, 22, 33) (2, 3, 12, 15, 60) (2, 3, 12, 16, 48) (2, 3, 12, 18, 36) (2, 3, 12, 20, 30) (2, 3, 12, 21, 28) (2, 3, 14, 15, 35) (2, 4, 5, 30, 60) (2, 4, 5, 36, 45) (2, 4, 6, 15, 60) (2, 4, 6, 16, 48) (2, 4, 6, 18, 36) (2, 4, 6, 20, 30) (2, 4, 6, 21, 28) (2, 4, 7, 12, 42) (2, 4, 7, 14, 28) (2, 4, 8, 10, 40) (2, 4, 8, 12, 24) (2, 4, 9, 12, 18) (2, 4, 10, 12, 15) (2, 5, 6, 9, 45) (2, 5, 6, 10, 30) (2, 5, 6, 12, 20) (3, 4, 5, 6, 20) –  Aug 31 '15 at 06:39
  • Note: The above sequences are generated using number less than 64 –  Aug 31 '15 at 07:11
  • Using numbers less then 200, I've got 114 possibilities for length 5, too many to list them here. –  Aug 31 '15 at 07:49
  • Correction: 57, not 114 –  Aug 31 '15 at 08:33
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    Because there are always only finitely many solutions to $\frac1{a_1}+\cdots+\frac1{a_n}\in\Bbb N$ we have that the given implication is an equivalence: $\sum_{s\in S}\frac1s\in\Bbb N\iff\sum_{s\in S}\frac1{f(s)}\in\Bbb N$. – Bart Michels Aug 31 '15 at 08:34
  • @ErickWong Looks like a promising idea. Do you know any lower bounds on the number of solutions to $\frac1{a_1}+\cdots+\frac1{a_n}\in\Bbb N$? It's not hard to think of some upper bounds but getting a lower bound seems harder to me. – Bart Michels Aug 31 '15 at 08:41
  • The source of the problem is : Romania TST 2013 –  Sep 06 '15 at 12:26

1 Answers1

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We have $f(1)=1$ and $f(n) \ge 2, \forall n \ge 2$.

Let $ n ≥ 2$ be un integer. Using Egyptian fractions theorem, we can write: $$ 1 − \frac {1}n = \sum_{s \in S} \frac{1}{s} $$ where $S$ is a set of integers greater than $n(n + 1)$. Therefore: $$ 1=\frac {1}n + \sum_{s \in S} \frac{1}{s} =\frac 1{n+1} + \frac 1{n(n+1)} + \sum_{s \in S} \frac{1}{s} $$ From f property, we have: $$ \frac {1}{f(n)} + \sum_{s \in S} \frac{1}{f(s)} \in \mathbb{N} $$ and $$ \frac 1{f(n+1)} + \frac 1{f(n(n+1))} + \sum_{s \in S} \frac{1}{f(s)} \in \mathbb{N} $$ therefore $$ \frac 1{f(n+1)} + \frac 1{f(n(n+1))} - \frac {1}{f(n)} \in \mathbb{Z} $$ But: $$ \frac {-1}2 \le - \frac {1}{f(n)} \lt \frac 1{f(n+1)} + \frac 1{f(n(n+1))} - \frac {1}{f(n)} \lt \frac 1{f(n+1)} + \frac 1{f(n(n+1))} \le \frac1{2} + \frac1{2} $$ so $$ \frac 1{f(n+1)} + \frac 1{f(n(n+1))} = \frac {1}{f(n)} \tag 1 $$ It follows that f is increasing and $f(n) \ge n$. To conclude, it's easy to show, using induction, that $f(n)=n, \forall n$.

Disclaimer

This prove has been sent to me, in a hand written form, by a friend who allowed me to post it here.

Update

I was requested to continue the prove (the induction part). First, because f is increasing and f injective, we have: $f(n) \ge n, \forall n$.

Now suppose $f(k) = k$ and $f(k + 1) > k + 1$ for some $k$. From (1) we have: $$ \frac 1{f(k+1)} + \frac 1{f(k(k+1))} = \frac {1}{k} \tag 2 $$ and, because $f(n) \ge n, \forall n$: $$ \frac 1{k+1} + \frac 1{k(k+1)} \gt \frac {1}{k} \tag 3 $$ From (3): $$ \frac {1}{k} \gt \frac {1}{k} \tag 4 $$ Therefore $f(k+1) = k+1$ if $f(k) = k$.

  • Something is wrong because $f(n)=1$ for all $n\in\mathbb{N}^*$ is a solution. – Batominovski Aug 31 '15 at 09:49
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    @Batominovski f is injective –  Aug 31 '15 at 09:55
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    Sorry, didn't see that you wanted "one-to-one" functions. However, I would be interested to see all solutions, not just injective ones. – Batominovski Aug 31 '15 at 09:56
  • @barto Right, thanks –  Aug 31 '15 at 10:07
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    I don't quite follow the very last step: the proof by induction. If you have the time, could you add this at the end of the question? Cheers. Otherwise, nice work. Kudos to your friend. – Colm Bhandal Sep 05 '15 at 20:44
  • Just to follow up, I can see it's obvious if $f$ is bijective, but I can't see how it follows from injectivity. – Colm Bhandal Sep 05 '15 at 21:01
  • @ColmBhandal It's not hard but I agree that it doesn't seem obvious enough to sweep under "easy to show, using induction". One approach is to make critical use of barto's comment that for any fixed $k$ there are only finitely many feasible $S$ with $|S|=k$. If $S_k$ is the union of all such $S$ then $f$ acts bijectively on $S_k$, and the result follows from $f(n)\ge n$ once we know that every positive integer belongs to some $S_k$. It's easier to use the stronger fact that $f$ is increasing, then it follows from the unboundedness of $\max S_k$. – Erick Wong Sep 06 '15 at 15:40
  • @ColmBhandal See the updated answer –  Sep 06 '15 at 17:48
  • @ErickWong See the updated answer –  Sep 06 '15 at 17:49