related to a famous olimpiad problem

harazi

Dear friends,
I know I posted many idiot questions during my presence on this site and this is the reason for which I will ask another one. Is there an infinite set of natural numbers such that the sum of the elements of any finite subset is a natural power? It is useless to say that I couldn't solve it, but all persons to whom I gave it didn' t manage to break it. Can you?

**Posted:** Sat Sep 11, 2004 5:08 am

heartwork

This seems to be related to Catalan's conjecture. If you have a proof for it, you might have an answer for this too!

**Posted:** Mon Sep 13, 2004 9:06 am

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Charlydif · Hodge Conjecture Offline Joined: 20 Aug 2003 Posts: 68

I'm not sure but i think the answer is yes. Someone told this problem related to the following:
Is there an infinite set of integers such that the sum of any finite subset its NOT a perfect power?
I remeber three things about this two problems,
1) One of them have a nice solution considering some density argument (probably the one with no sum perfect power)
2) One have a simple solution
3) The other was difficult and the person didnt told me the solution.

I am going to try to calrify my memory and think on them tomorrow, if i remeber or have any advance i post it.
For instance i think itsnt dificult to show there are arbitrary long sequences with this property but i am not sure right now. Anyway, i cant see the conection with catalan conjecture! And which one is the famous olympiad problem?

**Posted:** Thu Sep 23, 2004 6:05 pm

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pbornsztein

Well, to prove that there exists an infinite set of natural numbers such that the sum of any finite of its non-empty subsets is not a perfect power is not difficult using the chinese remainders theorem :

Start from $x_1 = 2$ .
Now, assume that we have defined $x_1, \ldots , x_n$ such that the sum of any non empty set whose elements are from the $x_i$ 's is not a power. We numerate these sets from $1$ to $k = 2^n-1$ . Let $s_i$ be the sum of the elements from the set numbered $i$ and $s_0 = 0$ .
Let $p_0,p_1, \ldots, p_k$ be pairwise distinct primes such that $s_i \neq 0$ mod [ $p_i$ ] for each $i>0$ (and $p_0$ is arbitrary)

From the chinese remainders theorem, there exists a positive integer $x$ such that $x = -s_i + p_i$ mod[ $p_i^2$ ] for each $i$ .
Thus $x+s_i$ is divisible by $p_i$ but not by $p_i^2$ . It follows that $x+s_i$ is not a power for all $i \geq 0$ . We then defined $x_{n+1} = x$ .

By induction, the above construction leads to an infinite set with the desired property.

But the question asked by Harazi seems more difficult.

Pierre.

**Posted:** Thu Sep 23, 2004 11:10 pm

harazi

Yes, that was the famous olympiad problem. I gave this new problem to my math teacher (who, believe me, is really a genius!) and he could not solve it. He gave it to other teachers and none could find a solution. Heartwork, I am also blind and I cannot figure out how it follows from Catalan conjecture. And I strongly think that such a set does not exist. Here is another beautiful question: is there an infinite set of natural numbers with the property that the sum of the inverses of the sum of the elements from all non-empty and finite subsets is convergent?

**Posted:** Fri Sep 24, 2004 9:30 am

Ravi B

**Posted:** Fri Sep 24, 2004 10:20 am

harazi

Hello! Does anyone know whether the following result has been proved or disproved:
Given a number $a$ and some pairs of numbers $m,n$ such that $min(m,n)<b$ where b is also a given number, the equation $x^m-y^n=a$ has only finitely many solutions in (x,y,m,n)? I imagine it is true, but too difficult to prove. Probably vess or Pierre know something? If the answer is yes, the problem is solved: there do not exists such a strange set.

**Posted:** Sat Apr 02, 2005 5:38 am

Bojan Basic

**Posted:** Fri Dec 09, 2005 4:57 pm

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- Why do mathematicians get Christmas and Halloween mixed up?
- Because Dec. 25 = Oct. 31.

harazi

Do you have the article? Would it be possible to attach it?

**Posted:** Sat Dec 10, 2005 2:19 am

Bojan Basic

I have the article, but I believe I can't attach it here because of copyright. However, if you give me your e-mail (by PM if you prefer) I would be glad to send the article.

**Posted:** Sat Dec 10, 2005 4:12 pm

_________________
- Why do mathematicians get Christmas and Halloween mixed up?
- Because Dec. 25 = Oct. 31.

seamusoboyle

**Posted:** Sat Dec 10, 2005 4:50 pm

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Newton is dead, Einstein is dead, and im not feeling too good myself...

Bojan Basic

**Posted:** Tue Jun 17, 2008 6:51 am

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- Why do mathematicians get Christmas and Halloween mixed up?
- Because Dec. 25 = Oct. 31.

Myth

It seems Thue equation involves terms of the same power only.
I believe finiteness of set of solutions $x^n-y^m=k$ is still an open question (at least my book of 1980 says so).

**Posted:** Tue Jun 17, 2008 10:59 am

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Myth is out of here

Bojan Basic

**Posted:** Tue Jun 17, 2008 12:25 pm

_________________
- Why do mathematicians get Christmas and Halloween mixed up?
- Because Dec. 25 = Oct. 31.