I was reading about A. Wiles solving Fermat's Last Theorem recently, and I would like to know any other conjectures or any other mathematical problem that has yet to be solved by anyone.

Does anyone know any?

Richard

The Twin Prime Conjecture There are an infinite number of twin primes, i.e. numbers $n$ where both $n$ and $n+2$ are prime (.e.g 11 and 13, 17 and 19, etc.)

You can find many more by going to http://mathworld.wolfram.com and typing "conjecture" in the search box.

Specifically, you can find some particularly important ones at this address

http://mathworld.wolfram.com/UnsolvedProblems.html

Thanks Dan

Richard

Here is another interesting example of an unsolved problem. (At least it was unsolved when I read an article about it 7 years ago.)

We have a sequence starting with any positive integer. To get to the next number in the sequence you either half the previous number (if it is even) or triple it and add one (if the last number is odd).

So here is an example starting with 10:

10 5 16 8 4 2 1 4 2 1 4 2 1 ...

and the 4 2 1 recurrs.

The theory is that no matter what the starting integer is the sequence always recurrs 4 2 1. But this had not yet been proven at the time of writing.

Michael

The first step to solving the above is to state that, because the 4 always follows one and 2 follows four, once you're to 1, 4 2 1 will reoccur. It has been proved that this sequance is true to a very high number. So this problem can be immediately shortened to whether every odd number can be multiplied by three, have 1 added, and be divided by four. If this is true, then the sequence is true. If it isn't though, this isn't necessarily wrong.

Just worked on this some, this statement simplifies to for every number 3/8n is an integer. This rule breaks at 1. Sorry if I got your hopes up.

But, this does show that every number 8n +1 works out.

Hi Brad!

This is a good first suggestion on how to go about this problem. However I'm not sure I follow your reasoning that 8n + 1 will always work out.

8n + 1 -> 24n + 4 -> -> 6n + 1 -> 18n + 4 -> 9n + 2 -> ???

Now we have a problem because the next stage depends on whether n is odd or even.

One approach I had considered was to try looking at the sequence in reverse. We start with one. Now for each successive number there are two options:

1) We can double the sequence.
2) We can subtract one and divide by 3.

However we are only allowed to perform 2) if the number in the sequence is 2 fewer than a multiple of 6 (can you see why this is necessary?).

So the question now is can our sequence reach all the integers if we play the moves appropriately?

(This approach was just an idea; it really hasn't got me anywhere.)

Looking at the problem generally, we can see that if the statement were false, there are two possibilities:

1) The sequence tends to infinity as the number of terms tends to infinity.

2) The sequence ends up oscillating

I'm not sure whether the number theorists have managed to eliminate at least one of these.

By the way, if the sequence starts at n and then goes up and down x times each alternately then you are left with:

(3/2)^x (n+1)-1

but this is of course a very special case. I'm going to see if I can generalise it...

Many thanks,

Michael

Yes, I once again checked my work, and it only extends to one step out. Never the less, if one could prove my above logic- that all odd numbers times three and plus one can be divided by four, the problem would be solved.

I doubt this helps; but oh, well,

Brad

If I understand you correctly, you are
saying that if n is odd, then 4 divides 3n+1.
Try n = 3.

Jonathan