
Theorems that are 'obvious' but hard to prove - ColinWright
http://mathoverflow.net/questions/51531/theorems-that-are-obvious-but-hard-to-prove
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lutusp
> Can anyone suggest other such theorems, in any areas of mathematics?

I vote for the conjecture of the infinity of Mersenne primes, the
"Lenstra–Pomerance–Wagstaff conjecture":

[http://en.wikipedia.org/wiki/Mersenne_conjectures#Lenstra.E2...](http://en.wikipedia.org/wiki/Mersenne_conjectures#Lenstra.E2.80.93Pomerance.E2.80.93Wagstaff_conjecture)

The topic is open, we're nowhere near outlining a proof, but their infinity
seems likely.

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ColinWright
So, they're asking for theorems (hence proven) that seem obvious but are hard
to prove, and you offer something that is not obvious, and unproven?

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lutusp
The Lenstra–Pomerance–Wagstaff conjecture seems obvious, and it's hard to
prove. What did I miss?

Given that there are an infinity of primes, then it seems obvious that an
infinite subset will be Mersenne primes. To dispute this, one would need to
imagine a mechanism that would prevent an infinite set of primes from
possessing an infinite subset of Mersenne primes.

But proving this is much harder than stating it.

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ColinWright

      > What did I miss?
    

They're asking for theorems, which means things that have been proven.

    
    
      > ... it seems obvious that an infinite subset
      > will be Mersenne primes.
    

Not to me - I can easily picture these things petering out after a time. For
example, it is suggested that there are only finitely many Fermat primes.

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lutusp
> They're asking for theorems, which means things that have been proven.

I managed to miss that. My bad.

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ColinWright
NP.

I, personally, also don't think it's obvious there are infinitely many
Mersenne primes.

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lutusp
> I, personally, also don't think it's obvious there are infinitely many
> Mersenne primes.

If I were a better mathematician, I might agree with you. But I can't think of
a compelling reason why some percentage of the infinity of primes would not
coincide with the infinity of members of (2^n)-1 , 1 <= n <= ∞, thereby
creating an infinite set.

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ColinWright
Then why not an infinite intersection between the primes and the numbers of
the form 2^(2^n)-1? And yet we only know 5 of them.

Let me help your intuition. Take these numbers:

    
    
      k(1) = 0
      k(2) = 2
      k(3) = 3
      k(n) = k(n-2)+k(n-3)
    

For a given _n_ , ask if _n_ divides _k(n)_ You'll find up to 10^5 or so this
works if and only if _n_ is prime. Does it work forever?

Consider the primes of the form 4k+1. For the first 10^5 or so of these they
turn out to the be sum of two squares. Does it work forever?

Heuristic arguments are all well and good, but at the end of the day, these
things need to be proven or disproven before we can be sure.

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lutusp
> ... at the end of the day, these things need to be proven or disproven
> before we can be sure.

Yes, and in the meantime, we're free to conjecture -- within reason.

