On Sunday, April 23, 2017 at 1:20:05 PM UTC-5, bassam

for

*Post by bassam king karzeddin**Post by bassam king karzeddin*a given Polynomial of (nth) degree, and becomes

easily clearer if the given polynomial of odd

degree,

first by

or

form,

*Post by bassam king karzeddin**Post by bassam king karzeddin*for a solution as (n/m), for some nonzero integer

(n),

And if your solution requires both your integers

(n

digits,

constructible

does

*Post by bassam king karzeddin**Post by bassam king karzeddin*not exist for that polynomial or that same

Diophantine equation

Regards

Bassam King Karzeddin

22th, April, 2017

It seems that I have to make all the work ready for

everyone professional independently and step by step

in order to make the idea clearer, despite providing

you all the scattered pieces essential in my posts to

complete this easy work for sure

*Post by bassam king karzeddin*so let us consider polynomials with rational

coefficients as the first step, and jumping on the

first and second degree polynomials, and choosing the

third degree polynomial (reduced form) to explain

this simple idea

*Post by bassam king karzeddin*(x^3 + ax + b = 0)

In any case a real (x) can be obtained as a

solution, and (x) is ultimately expressed as a

fraction in decimal representation (i.e rational

number), to some finite sequence of digits as a

rational solution or approximate solution once (x)

considered irrational number

*Post by bassam king karzeddin*So, let (x = n/m), and (a = a_1/a_2), and (b =

b_1/b_2), where all the terms (n, m, a_1, a_2, b_1,

b_2) are integers, and (m*(a_2)*(b_2) =/= 0)

*Post by bassam king karzeddin*Then we can simply get the following Diophantine

equation by direct substitution

*Post by bassam king karzeddin*(a_2)*(b_2)*n^3 + (a_1)*(b_2)*n*m^2 +

(a_2)*(b_1)*m^3 = 0

*Post by bassam king karzeddin*So, when does this integer equation have solution?

Oops, my time is out now, to be completed later

Any way it is easy beyond limits, also extending

this simplest idea is not any harder problem, but the

point is that all polynomials with rational

coefficients are actually and only Diophantine

equations for sure

*Post by bassam king karzeddin*Regards

Bassam King Karzeddin

23th, April, 2017

I'm obviously missing your point, but wouldn't the

rational root theorem allow you to avoid these lovely

Diophantine equations

Don

The rational root theorem is checking if a rational number is the solution expressed in simple form as (n/m), for some finite integers

But when there is not any real solution the alleged solution of cubic equation formula for the real solution pretends a fake solution again in the same simple form above (n/m), but in this case both integers are with infinite sequence of digits, where simply they take few digits as a solution to convince the poor student that was really a real solution

But the simplest obvious fact that such a solution can never exists, being unreal and non existing for sure, since such division of two integers where each consists of infinite sequence of digits is not permissible in the holy grail principle of mathematics nor defined plus also impossible task or achievement for sure

This is actually the true meaning of non solvability of some Diophantine equation

Then there is not any shame if the currents mathematics confesses the truth and simply say the obvious fact, that solution generally does not exist, but only approximation solution instead of committing three obvious crimes against human minds (this is the whole point)

Crime 1) permitting the division operation of two undefined integers where each of them must consists of unknown infinite sequence of digits

Crime 2) Convincing the student that solution exists at infinity, which is not true absolutely

Crime 3) Obtaining the solution again in rational or generally constructible form ( however there is no other way except meaningless notations in mind only) and convincing the innocents students that is irrational solution

We must realize also that the ONLY proved root operation is the SQUARE root operation by the Pythagoras theorem, whereas other higher root operations (not purely even) were never proved but wrongly concluded, check it from history please

One day (before more than two thousands years), it was a very big discovery and a real revolution in the mathematical science for irrational numbers by the Pythagoreans, as Sqrt(2), but we never heard of alike discovery for 2^{1/3}, or higher valid proved root operation and purely even

Regards

Bassam King Karzeddin

24th, April, 2017