On the road to prove Fermat's Impossiblity Equation !

[imatfaal]

 

I think that the equation indicates (4xy³ - x⁴)^1/3 rather than 3^1/2 multiplied by(4xy³ - x⁴)^1/2

 

It doesn't - but if that is what is required then so be it. Where is the equation from? And you do realise this is a fool's errand? Wiles showed that Fermat was correct - this would lead to a counter-example ...

 

is this the equation you were looking to represent Thomas?

 

[latex] z=\frac{\sqrt[3]{4xy^3-x^4}-3x^2+6x}{6x} [/latex]

[Commander]

 

It doesn't - but if that is what is required then so be it. Where is the equation from? And you do realise this is a fool's errand? Wiles showed that Fermat was correct - this would lead to a counter-example ...

 

is this the equation you were looking to represent Thomas?

 

[latex] z=\frac{\sqrt[3]{4xy^3-x^4}-3x^2+6x}{6x} [/latex]

 

Hi imatfaal,

 

Yes, Thank you.

 

That's the Equation and the last term 6x on the numerator can be dropped too.

 

Then z will become z -1 and that is OK too as an integer.

 

I am unable to draw it out correctly

 

Regards

Edited January 3, 2015 by Commander

[imatfaal]

Assuming x,y, z can be any integer and keep the LHS as "just any integer" [latex]\mathbb{Z}[/latex] stands for any integer

 

[latex] z=\frac{\sqrt[3]{4xy^3-x^4}-3x^2+6x}{6x} [/latex]

 

[latex] z - \frac{6x}{6x}=\frac{\sqrt[3]{4xy^3-x^4}-3x^2}{6x} [/latex]

 

[latex] \mathbb{Z} =\frac{\sqrt[3]{4xy^3-x^4}-3x^2}{6x} [/latex]

 

remembering that 6x must be an integer if x is an integer

 

[latex] \mathbb{Z} =\sqrt[3]{4xy^3-x^4}-3x^2 [/latex]

 

and that -3x^2 must be as well

 

[latex] \mathbb{Z} =\sqrt[3]{4xy^3-x^4}[/latex]

 

 

BUT - and it is a BIG BUT ; Andrew Wiles has already proven that this cannot be the case!!!!

[Commander]

Assuming x,y, z can be any integer and keep the LHS as "just any integer" [latex]\mathbb{Z}[/latex] stands for any integer

 

[latex] z=\frac{\sqrt[3]{4xy^3-x^4}-3x^2+6x}{6x} [/latex]

 

[latex] z - \frac{6x}{6x}=\frac{\sqrt[3]{4xy^3-x^4}-3x^2}{6x} [/latex]

 

[latex] \mathbb{Z} =\frac{\sqrt[3]{4xy^3-x^4}-3x^2}{6x} [/latex]

 

remembering that 6x must be an integer if x is an integer

 

[latex] \mathbb{Z} =\sqrt[3]{4xy^3-x^4}-3x^2 [/latex]

 

and that -3x^2 must be as well

 

[latex] \mathbb{Z} =\sqrt[3]{4xy^3-x^4}[/latex]

 

 

BUT - and it is a BIG BUT ; Andrew Wiles has already proven that this cannot be the case!!!!

 

Hi imatfaal and Michel123456,

 

I am very sorry.

 

After I checked again with the help of the tool it appears my interpretation was wrong.

 

So it is only

 

z = [ 3^1/2 (3xy³ - x⁴) ^1/2 - 3x² ] / 6x

 

and therefore we need to contend with Square root of 3 which is an irrational number !

 

Because this is the Solution for z in the Equation y³ + z³ = (z+x)³ or y³ = (z+x)³ - z³ as a Sum or difference between Cubic Numbers and therefore the solution to y³ + z³ = n³ has only an irrational Value and hence proves the Fermat's Conjecture for the Power of 3.

 

 

 

Edited January 3, 2015 by Commander

[michel123456]

My "feeling" with Fermat's last theorem is that you need an equal number of factors for having integers in the solution.

For example a³ + b³ = c³ will not give integer but a³ + b³ + c³= d³ will.

And

a⁴ + b⁴ + c⁴+ d⁴= e⁴ will have an integer solution.

 

But that is just a feeling.

[Commander]

My "feeling" with Fermat's last theorem is that you need an equal number of factors for having integers in the solution.

For example a³ + b³ = c³ will not give integer but a³ + b³ + c³= d³ will.

And

a⁴ + b⁴ + c⁴+ d⁴= e⁴ will have an integer solution.

 

But that is just a feeling.

 

Yes, 3³ + 4³ + 5³ = 6³

 

Many Solutions for a³ + b³ = c² might be found I think. like 1³ + 2³ = 3^{2 ,}

 

It is also very interesting to note that [1³+2³+3^{3 ...} + n³] = [1+2+3 .....+n]² = [n (n+1) /2] ²

 

 

Edited January 3, 2015 by Commander

[Commander]

I need to put some more info already known to me.

 

Particularly that z should be an irrational number whatever be the substitutions elsewhere.

 

Also efforts will still be on for more Solutions

 

Pl give me some time.

 

PS : Overnight had a severe Tooth Ache which God has lessened it and now I have a high shivering too.

 

Com will be off for a while

i came in to quickly announce [it is already quite late] that it is only due to the Support and Discussion with imatfaal and michel123456 who pointed out the errors which I was making so that I could Correct the equation !

 

Thank you Friends !

 

>>>>>>>>>>>

 

Another thing I can add here is :

 

To produce a Cubic Number C_n the Sum W_{a [1 ,,, n) can also be used with} W_{a = 3a² - 3a +1 instead 0f (a = 0,, n-1)}

_{and both will be the same !!}

Edited January 4, 2015 by Commander

[Sensei]

Any Positive Natural Number N can be written as a positive sum of the Powers of 2 !

 

N = 2^a + 2^b ..... with as many terms as require where a , b , c etc are all Natural positive Numbers from [ 0,1,2,3 .....]

 

These powers a, b, etc in the sequence will be PRESENT or ABSENT in the Equation only once.

 

No repetition is required.

 

This is nothing but a Binary Number representing the powers of 2 which add up to be equal to N.

 

Just a small note:

We can have binary number that's not integer, but finite fractional.

f.e. if we write:

0.1011

it will be:

2^-1+2^-3+2^-4=0.5+0.125+0.0625=0.6875

 

Binary %1011 is 11 in base 10.

11/16=0.6875

Edited January 25, 2015 by Sensei

[Commander]

 

Just a small note:

We can have binary number that's not integer, but finite fractional.

f.e. if we write:

0.1011

it will be:

2^-1+2^-3+2^-4=0.5+0.125+0.0625=0.6875

 

Binary %1011 is 11 in base 10.

11/16=0.6875

Yes, Sensei, but here I was talking only about Integers