In article ,
James Harris wrote:
(Arturo Magidin) wrote in message ...
In article ,
James Harris wrote:
(Arturo Magidin) wrote in message ...
In article ,
James Harris wrote:

[.snip.]

It's not possible. I'll put in values to help those readers confused
by symbols (though algebra is BASED on symbols) by letting u=2, f=13,
then you have

26/w_1(m)

and you don't have to be a rocket scientist to know that no function
w_1(m) that actually varies with m can exist such that 26/w_1(m) is an
algebraic integer for *all* integer m.


w_1(m) = 13^{1/(m^2+1)}*2^{|m|/(m^2+1)}

I'm not saying that's the function in question (it is not), but there
is a function that "actually varies with m such that 26/w_1(m) is an
algebraic integer for *all* integer m." So the argument that no such
function can exist is simply bogus.

Well that example works, so now I'll say algebraic integer m, and it
blows up at m^2+1 = 0.

So let me see if I have this right:

(1) You claimed something was impossible under a given set of
conditions.

(2) When I proved that your claim was simply false, you changed the
set of conditions, and now argue that what I said was wrong.

You ****ing dumbass Arturo Magidin, the ring has ALWAYS been algebraic
integers, so having m in the ring of algebraic integers FOLLOWS ANYWAY
you goddamn, ****ing, stupid dumbass.


You asked for a function that satisfied certain, specific, explicitly
given properties, claiming such a function was impossible.

I gave a function that satisfied EACH AND EVERY ONE of the properties
you listed.

Then you changed the properties you wanted. I noted the change, and
proceeded to give you a function that satisified all the NEW
properties, AND MORE (being defined over all the algebraic numbers, not
just the algebraic integers; you can restrict it to the algebraic
integers if you want). But you deleted it. Here it is again:

-- Begin Insert --

Here's an example that works for EVERY algebraic NUMBER m:

Step 1. Given an algebraic number m, let f(x) be the unique monic
polynomial with rational coefficients, irreducible over Q, which has m
as a root. Write it as:

f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0.

Note that a_0 must be different from 0.

Step 2. Write a_0, a rational number, as a_0 = r/s, with r and s
integers, r and s relatively prime.

Step 3. If r=1 or -1, let q = 1.

Step 4. If r is not equal to 1 or -1, then let q be the largest
rational prime that divides r.

Step 5. Let w(m) be a root of the polynomial

x^4 + 13qx^3 + qx + 13.


Then:

(a) w(m) is an algebraic integer.

(b) w(m) divides 13 in the ring of algebraic integers, since the
product of all the roots is 13, and every root is an algebraic
integer.

(c) w(m) is not constant: it takes different values at different
integers.

(d) w(m) takes each value a countably infinite number of times, so
w(m) cannot be given by a polynomial.


-- End Insert --

I gave a variant elsewhere defined for all complex numbers m as well.

Still think such a function is impossible?

================================================== ====================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A great
many people are staggered to this extent, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
================================================== ====================

Arturo Magidin