P(x) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) =
= (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22)
with
b_3(x) = a_3(x) - 3
using this substitution we get the factorisation:
= (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7)
where he says that the a's are roots of:
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
so far so good.

Oh well. James started one of his latest post with a polynomial with
factorisation (citations are not exact):
P(x) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) =
= (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22)
with
b_3(x) = a_3(x) - 3
using this substitution we get the factorisation:
= (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7)
where he says that the a's are roots of:
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
so far so good.

A lot of conclusions are drawn from this factorisation, it is however
the consequence of another factorisation. Set y = 49x, we get:
P(y) = 125 y^3 - 375 y^2 - 360 y + 1078 =
= (5 a_1(y) + 7)(5 a_2(y) + 7)(5 a_3(y) + 7)
where the a's are roots of:
a^3 + 3(-1 + y)a^2 - (y^3 - 3y^2 + 3y)

So although the "constant term" of two of the factors are divisible by
7, and the third is co-prime to it (it is 22), in general *none* of the
factors are divisible by 7, as P(y) is only divisible by 7 in a limited
number of cases. So I am still wondering what JSH is trying to show
with his "constant terms".

In sci.math, Dik T. Winter
Dik.Winter@cwi.nl
wrote
on Fri, 26 Dec 2003 03:06:36 GMT
HqHFB0.CvM@cwi.nl>:
....
A lot of conclusions are drawn from this factorisation, it is however
the consequence of another factorisation. Set y = 49x, we get:
P(y) = 125 y^3 - 375 y^2 - 360 y + 1078 =
= (5 a_1(y) + 7)(5 a_2(y) + 7)(5 a_3(y) + 7)
where the a's are roots of:
a^3 + 3(-1 + y)a^2 - (y^3 - 3y^2 + 3y)

So although the "constant term" of two of the factors are divisible by
7, and the third is co-prime to it (it is 22), in general *none* of the
factors are divisible by 7, as P(y) is only divisible by 7 in a limited
number of cases. So I am still wondering what JSH is trying to show
with his "constant terms".

So am I. I'd be curious to figure out when a_1(x)/7 and a_2(x)/7
are algebraic integers, but admittedly it's not a priority.

It's
clear that generally speaking, they are not, although
a_1(0) = 0, a_2(0) = 0, a_3(0) = 3.

Your manipulation is interesting and simplifies the problem
considerably. However, now one has to deal with y being
a multiple of 49, if x was originally an algebraic integer.

(I'm also curious as to the rest of his proof; this is
only a small snippet thereof. It's a bit like examining
a small area (sans the actual hole) of a flat tire to try
to figure out why the car won't move.)

In article <fk4ub1-8mh.ln1@lexi2.athghost7038suus.net> The Ghost In The Machine <ewill@sirius.athghost7038suus.net> writes:
In sci.math, Dik T. Winter
Dik.Winter@cwi.nl
wrote
on Fri, 26 Dec 2003 03:06:36 GMT
HqHFB0.CvM@cwi.nl>:
...
A lot of conclusions are drawn from this factorisation, it is however
the consequence of another factorisation. Set y = 49x, we get:
P(y) = 125 y^3 - 375 y^2 - 360 y + 1078 =
= (5 a_1(y) + 7)(5 a_2(y) + 7)(5 a_3(y) + 7)
where the a's are roots of:
a^3 + 3(-1 + y)a^2 - (y^3 - 3y^2 + 3y)

So although the "constant term" of two of the factors are divisible by
7, and the third is co-prime to it (it is 22), in general *none* of the
factors are divisible by 7, as P(y) is only divisible by 7 in a limited
number of cases. So I am still wondering what JSH is trying to show
with his "constant terms".

So am I. I'd be curious to figure out when a_1(x)/7 and a_2(x)/7
are algebraic integers, but admittedly it's not a priority.

You first are required to show what a_1, a_2 and a_3 are. Luckily that
is possible... (*)

It's
clear that generally speaking, they are not, although
a_1(0) = 0, a_2(0) = 0, a_3(0) = 3.

Your manipulation is interesting and simplifies the problem
considerably. However, now one has to deal with y being
a multiple of 49, if x was originally an algebraic integer.

But it also shows that the factors are not divisible by 7 for *all* y,
so it shows that divisibility properties are erratic.

(I'm also curious as to the rest of his proof; this is
only a small snippet thereof. It's a bit like examining
a small area (sans the actual hole) of a flat tire to try
to figure out why the car won't move.)

The rest of his proof hinges on the fact that exactly two factors are
(FLT proof) or should be (definition error) divisible by 7.

----
(*) A very nice showing of that I found while looking around at the
solutions of cubic equations. All expositions I have seen miss a very
basic fact (I think), but the exposition at mathworld is closest (this
is also inspired by an article by C. Bond).

Let's calculate the roots of z^3 + a.z^2 + b.z + c = 0. Let us have
the following definitions:
Q = (12.b - 4.a^2)
R = 36.a.b - 108.c - 8.a^3
K1 = cbrt(R + sqrt(Q^3 + R^2))/2
K2 = cbrt(R - sqrt(Q^3 + R^2))/2
W = (-1 + i.sqrt(3))/2, W^2 = (-1 -i.sqrt(3))/2, W^3 = 1
then we have the following roots:
z_1 = (-a + W .K1 + W^2.K2)/3
z_2 = (-a + W^2.K1 + W .K2)/3
z_3 = (-a + K1 + K2)/3
filling in all stuff (and hoping I did not make a basic arithmetic
mistace), we find that the z's correspondend to James' a's in order.

The beauty of this presentation is (I think) how three cubic roots
of two different values are used.