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.

Edited October 14, 20169 yr by Joatmon

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

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.

Well then there is your proof

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'.

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'.

 

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 October 16, 20169 yr by renerpho

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.

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.

 

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