Math Is Fun Forum

tony123

Member

Registered: 2007-08-03

Posts: 229

Find all pairs

Find all pairs of natural numbers (x,y) for which

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Find all pairs

I haven't managed to solve it yet, but maybe someone else can use what I've found so far.

First, I've found 3 solutions: (1,1), (2,1), and (4,2).  Secondly, x+1 must be prime.  This can be seen with a simple rearrangement:

x! + 1 cannot be divided by any number from 2 to x, which means (x+1)^y can't either, which means x+1 cannot be divided by any number from 2 to x, which makes it prime.

After this I'm stuck.  What's interesting is that there are solutions where x+1 is prime for each of the first 3 primes but none of the following 3, which is where I stopped trying to find solutions by hand.

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ZHero

Real Member

Registered: 2008-06-08

Posts: 1,889

Re: Find all pairs

What I have found out is that x and y should both be even numbers!
I'm working it out further though.. I guess it also makes use of "all powers of a given prime number in x!". I guess, if we can equate the powers of 2 (as both x and y are even) in LHS and RHS, we're done!

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ZHero

Real Member

Registered: 2008-06-08

Posts: 1,889

Re: Find all pairs

Oh! I forgot to state that x & y should be even except when one of them is 1 !!

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If two or more thoughts intersect, there has to be a point!

ZHero

Real Member

Registered: 2008-06-08

Posts: 1,889

Re: Find all pairs

There's one more minute thing i've found.. The number of 2s in (2^n)! is 2^n - 1.
But i don't see how to use it...

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If two or more thoughts intersect, there has to be a point!

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Find all pairs

Due to the fact that x+1 must be prime, which I showed above, it's clear that x must be even (or 1, since 1+1 is the only even prime number).  However, I can't see the why y must also be prime.  Could you please show us the justification for this?

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Find all pairs

However, I can't see the why y must also be prime.  Could you please show us the justification for this?

ZHero said y must be even.

I have a sneaking suspicion that this problem is combinatorial and not number theoretic, as you guys have been treating it.  But then again, I have never seen a combinatorial problem like this.  Unfortunately being at work means I can't really work in it anymore than that.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."