Sign in to follow this  
Joatmon

Solving a formula

Recommended Posts

I have reason to think that if A, B and N in the following equation are positive whole numbers then there is no solution. I would like either values of A, B and N which satisfy the equation or a reason why there is no solution. If you think it looks easy then give it a go - please.

post-68560-0-05643900-1476471681_thumb.jpg

Edited by Joatmon

Share this post


Link to post
Share on other sites

I am pretty sure not. But cannot yet prove it.

 

If you rearrange for a I think you get

 

[latex]a =\frac{ (\sqrt3 \sqrt{b (3 b^3+4 n^3)}-3 b^2+6 b n)}{(6 b)}[/latex]

 

If b and n are +ve integers then it is obvious that every term except

 

[latex] \sqrt3 \sqrt{b (3 b^3+4 n^3)}[/latex]

 

is an integer. So for a to be an integer then that term must also be an integer (ie if it is less than 1 then no matter how much you add to it - or whatever integer you divide it by it will always be less than 1. or if it is greater than 1 then it cannot be irrational for similar reasons). I feel I should be able to show that cannot be the case - but I cannot

Share this post


Link to post
Share on other sites

I'm not surprised - if I have worked things out correctly then a whole number solution would lead to a whole number example of A cubed plus B cubed = C cubed. If it can be proved that there is no whole number solution then it can be concluded that there is no whole number solution to A cubed plus B cubed = C cubed.

Share this post


Link to post
Share on other sites

Well then there is your proof

I think that sometimes a journey and its byways can be as interesting as a destination never reached. Also, I feel certain that there are a number of people in the world, such as me, who although knowing they are almost certainly deluded, think that there is just a possibility that there is still to be found a much more simple way of proving Fermat's Last Theorem than that produced by Andrew Wiles'. :D

Share this post


Link to post
Share on other sites

I think that sometimes a journey and its byways can be as interesting as a destination never reached. Also, I feel certain that there are a number of people in the world, such as me, who although knowing they are almost certainly deluded, think that there is just a possibility that there is still to be found a much more simple way of proving Fermat's Last Theorem than that produced by Andrew Wiles'. :D

 

Perhaps - but notice that, even if your formula is related to Fermat's Last Theorem for the case [math]n=3[/math], it can not attack the general case. And there already are easy and elementary proofs for the case [math]n=3[/math].

Edited by renerpho

Share this post


Link to post
Share on other sites

That is true and also for other specific values of n. However, perhaps the source of the formula I gave might lead to a generalisation for all values of n. I admit that is unlikely but it is an interesting conundrum.

Share this post


Link to post
Share on other sites

That is true and also for other specific values of n. However, perhaps the source of the formula I gave might lead to a generalisation for all values of n. I admit that is unlikely but it is an interesting conundrum.

 

Absence of evidence is not absolute evidence of absence; but when the greatest mathematicians search fruitlessly for a simple solution for centuries I think that - on the balance of probabilities - one can assume that a simple solution does not exist.

Share this post


Link to post
Share on other sites

 

Absence of evidence is not absolute evidence of absence; but when the greatest mathematicians search fruitlessly for a simple solution for centuries I think that - on the balance of probabilities - one can assume that a simple solution does not exist.

Can't argue with that :)

I guess that if I got a chance to play chess against a grand master I would jump at the chance and come back for more after losing. I suppose its a question of what one finds interesting or fun :D

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this