Saturday, April 26, 2003
Non algebraic integer "integer like" number, demonstrated
Well I use very simple ideas to try and get nifty answers to some old math problems. And I post about them. However, I am not a mathematician so at times I find myself in territory where I can be bullied a bit by mathematicians making bold claims that I'm wrong, and I have had that problem with a key area which has to do with the
coverage of algebraic integers.
Several posters have argued with me for some time about a relatively simple result that I have. The arguments have gone back and forth, and for some odd reason people don't seem to believe me!
However, looking carefully over the information posted by a guy who calls himself "Fred the Wonder Worm" I noticed that it demonstrated that the people arguing with me were wrong, and revealed an algebraic integer unit, which has a multiplicative inverse which is itself not an algebraic integer.
The following is taken from his post
message id: <b82u36$5aq$1@spacebar.ucc.usyd.edu.au>
I'll keep his letters to lessen confusion, and first notice from
x^12 + 6*x^11 - 181*x^10 - 960*x^9 + 7963*x^8 + 37678*x^7 - 117533*x^6 -
488570*x^5 + 686101*x^4 + 2232396*x^3 - 1726811*x^2 - 2931010*x + 987100
that taking a root, using 'a' as he did, and dividing through by 'a' gives
a^11 + 6*a^10 - 181*a^9 - 960*a^8 + 7963*a^7 + 37678*a^6 - 117533*a^5 -
488570*a^4 + 686101*a^3 + 2232396*a^2 - 1726811*a - 2931010 + 987100/a=0
where you can see that if 'a' is a root that has non unit algebraic integer factors of 5, then 987100/a, must STILL have non unit algebraic integer factors of 5, which means that a/5 does not.
For instance, if 'a' has a factors of 5sqrt(5), then because 987100 has a factor of 25 that would leave only sqrt(5), but a full factor of 5 has to still divide across because of 2931010 and all the other terms with a's, so there's be a contradiction.
Now looking at r1, from Fred's post I have
Putting in tr I have
176866667462880 r1 = 9862453210*a^11 + 54243492655*a^10 -
1198409518887700 - 1762144506621*a^9 - 8336476474707*a^8 + 73634871906622*a^7 +
297005313003728*a^6 - 915848551626666*a^5 - 3051857657480243*a^4 +
3027640866778493*a^3 + 7747460630371967*a^2 + 403832110132962*a
and dividing through by 5, as well as collecting a bit gives
35373333492576 r1 = 1972490642*a^11 + 10848698531*a^10 - 239681903777540 -
a/5( 1762144506621*a^8 + 8336476474707*a^7 - 73634871906622*a^6 -
297005313003728*a^5 + 915848551626666*a^4 + 3051857657480243*a^3 -
3027640866778493*a^2 - 7747460630371967*a + 403832110132962)
Let k = 1762144506621*a^8 + 8336476474707*a^7 - 73634871906622*a^6 -
297005313003728*a^5 + 915848551626666*a^4 + 3051857657480243*a^3 -
3027640866778493*a^2 - 7747460630371967*a + 403832110132962
and note that if 'a' has factors of 5, that k does not because
403832110132962
is, of course, NOT divisible by 5.
So I have
35373333492576 r1 = 1972490642*a^11 + 10848698531*a^10 - 239681903777540 - ak/5
Now then, if r1 shares any non unit algebraic integer factors of 5 with 'a', then they can divide across because of
- 239681903777540
which, of course, has a factor of 5.
Then ak/5 must also have those factors
But as I've shown above ak/5 cannot have non unit algebraic integer factors of 5, which forces r1 to be a unit, as I quote from Fred:
What makes that fascinating is that if you invert the polynomial using y=1/x, you get
5x^5 + 4176 x^3 - 1488 y^2 + 1
which has now been proven to have a unit root, which is, NOT an algebraic integer, but its inverse IS an algebraic integer.
That settles the argument I've had for several years now.
I think the math is nifty.
Here's the background if you're wondering.
I demonstrated recently a key result using the polynomial-like expression
(v^3+1)z^6 - 3v x^2 y^2 z^2 + x^3 y^3,
with v=-1 + mf^{2j}, where m=1, j=1, with the factorization
(v^3+1)z^6 - 3v x^2 y^2 z^2 + x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
that ONLY TWO of the a's could have a factor of f.
To make that simple enough, I showed f=0, which I'll give here quickly as it is
3 x^2 y^2 z^2 + x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
and it's easy enough to see that ONLY two of the a's are 0.
Then I went to f=sqrt(2), which gives the polynomial
2z^6 - 3 x^2 y^2 z^2 + x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
where it's easy to demonstrate that ONLY TWO of the a's have a factor of sqrt(2), but then I went to f=sqrt(5), and ran into a problem.
The problem is in demonstrating the factorization, but I stumbled upon a route that worked by accident.
First off the polynomial for f=sqrt(5) is
65 z^6 - 12 x^2 y^2 z^2 + x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
but because it's irreducible there's a twist because there have been posters arguing with me for over a YEAR that because it's irreducible, now ALL the a's would have non unit algebraic integer factors of 5. (Don't get confused on the usage here as it's like 3 is a factor of 12, so if I talk about factors of 12, I don't have to mean 12 itself.)
Well I have a VERY simple argument that proves these people wrong, but the arguing continued.
Now one of them had demonstrated a technique to isolate factors of 5, by solving for the a's first, which because b1 b2 b3=1 is easy here as I'd have
u^3 - 12 u^2 + 65 = 0
and the negative of the roots gives the a's, and using the substitution sqrt(5)w, I have
5sqrt(5) w^3 - 12(5) w^2 + 65 = 0, and dividing off 5, I get
sqrt(5) w^3 - 12 w^2 + 13 = 0, so
sqrt(5) w^3 = 12 w^2 - 13
5w^6 = 144 w^4 - 312 w^2 + 169, so
5w^6 - 144 w^4 + 312 w^2 - 169 =0
and the assertion by these people that argue with me is that because that polynomial is non monic primitive and irreducible over Q, none of the I know that the a's from before must all have factors of 5.
Now I erroneously believed that if they were right the factorization
5w^6 - 144 w^4 + 312 w^2 - 169 = (c1 w + d1)…(c6 w + d6)
where the c's and d's are algebraic integers could not be found.
Well two people responded by posting the polynomials defining the c's and d's where they were clearly algebraic integers, and I did a cursory check but haven't seen anything wrong with their factorization.
coverage of algebraic integers.
Several posters have argued with me for some time about a relatively simple result that I have. The arguments have gone back and forth, and for some odd reason people don't seem to believe me!
However, looking carefully over the information posted by a guy who calls himself "Fred the Wonder Worm" I noticed that it demonstrated that the people arguing with me were wrong, and revealed an algebraic integer unit, which has a multiplicative inverse which is itself not an algebraic integer.
The following is taken from his post
message id: <b82u36$5aq$1@spacebar.ucc.usyd.edu.au>
Let a be a root of this polynomial:
x^12 + 6*x^11 - 181*x^10 - 960*x^9 + 7963*x^8 + 37678*x^7 - 117533*x^6 -
488570*x^5 + 686101*x^4 + 2232396*x^3 - 1726811*x^2 - 2931010*x + 987100
Define tr, r1, r2, r3 to be the following values:
tr = 176866667462880
r1 = (9862453210*a^11 + 54243492655*a^10 - 1762144506621*a^9 -
8336476474707*a^8 + 73634871906622*a^7 + 297005313003728*a^6 -
915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +
7747460630371967*a^2 + 403832110132962*a - 1198409518887700) / tr
r2 = (6573801085*a^11 + 29503072348*a^10 - 1208403339336*a^9 -
4366240041747*a^8 + 54222010610962*a^7 + 148750611762872*a^6 -
793269747747321*a^5 - 1487188701233408*a^4 + 3985921497989948*a^3 +
4839390026896655*a^2 - 5550606915303438*a - 5671414313897740) / tr
r3 = (6573801085*a^11 + 42808739587*a^10 - 1141875003141*a^9 -
6752351088912*a^8 + 44278396405132*a^7 + 248393754086744*a^6 -
459258252043281*a^5 - 2663297151231683*a^4 + 1065284831907473*a^3 +
7093998365144384*a^2 - 1778334433861968*a - 130135870547080) / tr
Note that r1, r2, r3 are all algebraic integers, as they are roots of
the (irreducible) monic polynomial x^6 - 1488*x^4 + 4176*x^2 - 5.
I'll keep his letters to lessen confusion, and first notice from
x^12 + 6*x^11 - 181*x^10 - 960*x^9 + 7963*x^8 + 37678*x^7 - 117533*x^6 -
488570*x^5 + 686101*x^4 + 2232396*x^3 - 1726811*x^2 - 2931010*x + 987100
that taking a root, using 'a' as he did, and dividing through by 'a' gives
a^11 + 6*a^10 - 181*a^9 - 960*a^8 + 7963*a^7 + 37678*a^6 - 117533*a^5 -
488570*a^4 + 686101*a^3 + 2232396*a^2 - 1726811*a - 2931010 + 987100/a=0
where you can see that if 'a' is a root that has non unit algebraic integer factors of 5, then 987100/a, must STILL have non unit algebraic integer factors of 5, which means that a/5 does not.
For instance, if 'a' has a factors of 5sqrt(5), then because 987100 has a factor of 25 that would leave only sqrt(5), but a full factor of 5 has to still divide across because of 2931010 and all the other terms with a's, so there's be a contradiction.
Now looking at r1, from Fred's post I have
tr = 176866667462880
r1 = (9862453210*a^11 + 54243492655*a^10 - 1762144506621*a^9 -
8336476474707*a^8 + 73634871906622*a^7 + 297005313003728*a^6 -
915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +
7747460630371967*a^2 + 403832110132962*a - 1198409518887700) / tr
Putting in tr I have
176866667462880 r1 = 9862453210*a^11 + 54243492655*a^10 -
1198409518887700 - 1762144506621*a^9 - 8336476474707*a^8 + 73634871906622*a^7 +
297005313003728*a^6 - 915848551626666*a^5 - 3051857657480243*a^4 +
3027640866778493*a^3 + 7747460630371967*a^2 + 403832110132962*a
and dividing through by 5, as well as collecting a bit gives
35373333492576 r1 = 1972490642*a^11 + 10848698531*a^10 - 239681903777540 -
a/5( 1762144506621*a^8 + 8336476474707*a^7 - 73634871906622*a^6 -
297005313003728*a^5 + 915848551626666*a^4 + 3051857657480243*a^3 -
3027640866778493*a^2 - 7747460630371967*a + 403832110132962)
Let k = 1762144506621*a^8 + 8336476474707*a^7 - 73634871906622*a^6 -
297005313003728*a^5 + 915848551626666*a^4 + 3051857657480243*a^3 -
3027640866778493*a^2 - 7747460630371967*a + 403832110132962
and note that if 'a' has factors of 5, that k does not because
403832110132962
is, of course, NOT divisible by 5.
So I have
35373333492576 r1 = 1972490642*a^11 + 10848698531*a^10 - 239681903777540 - ak/5
Now then, if r1 shares any non unit algebraic integer factors of 5 with 'a', then they can divide across because of
- 239681903777540
which, of course, has a factor of 5.
Then ak/5 must also have those factors
But as I've shown above ak/5 cannot have non unit algebraic integer factors of 5, which forces r1 to be a unit, as I quote from Fred:
Note that r1, r2, r3 are all algebraic integers, as they are roots of the (irreducible) monic polynomial x^6 - 1488*x^4 + 4176*x^2 - 5.
What makes that fascinating is that if you invert the polynomial using y=1/x, you get
5x^5 + 4176 x^3 - 1488 y^2 + 1
which has now been proven to have a unit root, which is, NOT an algebraic integer, but its inverse IS an algebraic integer.
That settles the argument I've had for several years now.
I think the math is nifty.
Here's the background if you're wondering.
I demonstrated recently a key result using the polynomial-like expression
(v^3+1)z^6 - 3v x^2 y^2 z^2 + x^3 y^3,
with v=-1 + mf^{2j}, where m=1, j=1, with the factorization
(v^3+1)z^6 - 3v x^2 y^2 z^2 + x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
that ONLY TWO of the a's could have a factor of f.
To make that simple enough, I showed f=0, which I'll give here quickly as it is
3 x^2 y^2 z^2 + x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
and it's easy enough to see that ONLY two of the a's are 0.
Then I went to f=sqrt(2), which gives the polynomial
2z^6 - 3 x^2 y^2 z^2 + x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
where it's easy to demonstrate that ONLY TWO of the a's have a factor of sqrt(2), but then I went to f=sqrt(5), and ran into a problem.
The problem is in demonstrating the factorization, but I stumbled upon a route that worked by accident.
First off the polynomial for f=sqrt(5) is
65 z^6 - 12 x^2 y^2 z^2 + x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
but because it's irreducible there's a twist because there have been posters arguing with me for over a YEAR that because it's irreducible, now ALL the a's would have non unit algebraic integer factors of 5. (Don't get confused on the usage here as it's like 3 is a factor of 12, so if I talk about factors of 12, I don't have to mean 12 itself.)
Well I have a VERY simple argument that proves these people wrong, but the arguing continued.
Now one of them had demonstrated a technique to isolate factors of 5, by solving for the a's first, which because b1 b2 b3=1 is easy here as I'd have
u^3 - 12 u^2 + 65 = 0
and the negative of the roots gives the a's, and using the substitution sqrt(5)w, I have
5sqrt(5) w^3 - 12(5) w^2 + 65 = 0, and dividing off 5, I get
sqrt(5) w^3 - 12 w^2 + 13 = 0, so
sqrt(5) w^3 = 12 w^2 - 13
5w^6 = 144 w^4 - 312 w^2 + 169, so
5w^6 - 144 w^4 + 312 w^2 - 169 =0
and the assertion by these people that argue with me is that because that polynomial is non monic primitive and irreducible over Q, none of the I know that the a's from before must all have factors of 5.
Now I erroneously believed that if they were right the factorization
5w^6 - 144 w^4 + 312 w^2 - 169 = (c1 w + d1)…(c6 w + d6)
where the c's and d's are algebraic integers could not be found.
Well two people responded by posting the polynomials defining the c's and d's where they were clearly algebraic integers, and I did a cursory check but haven't seen anything wrong with their factorization.
# posted by James Harris @ 17:49 
