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Old 2017-04-07, 00:35   #1
carpetpool
 
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Post Polynomial whose coefficients add up to n defining Cyclotomic field K.

Let n be an integer defining the cyclotomic properties of K (meaning that n is a factor of the cyclotomic polynomial C_K(x) evaluated at some x value). How many polynomials P(x), the same degree as C_K(x) their coefficients add up to n? For instance, choosing the cyclotomic field 3, C_3(x) = x^2+x+1, and x = 7, 19 is a factor of 7^2+7+1. How many polynomials P(x) of the form ax^2+bx+c defining the same field as x^2+x+1 is it the case that a+b+c = 19 where a, b, c are integers -19 <= (a, b, c) <= 19? Thanks for help, comments, and clarification.

Last fiddled with by carpetpool on 2017-04-07 at 00:35
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Old 2017-04-07, 00:50   #2
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Quote:
Originally Posted by carpetpool View Post
Let n be an integer defining the cyclotomic properties of K (meaning that n is a factor of the cyclotomic polynomial C_K(x) evaluated at some x value). How many polynomials P(x), the same degree as C_K(x) their coefficients add up to n? For instance, choosing the cyclotomic field 3, C_3(x) = x^2+x+1, and x = 7, 19 is a factor of 7^2+7+1. How many polynomials P(x) of the form ax^2+bx+c defining the same field as x^2+x+1 is it the case that a+b+c = 19 where a, b, c are integers -19 <= (a, b, c) <= 19? Thanks for help, comments, and clarification.
well the number of polynomials total before the field consideration is 6*partitions(19,,[3,3]) ( as there are 6 orders possible for {a,b,c} ) or 180 ( okay I see now, you include negatives which throw the numbers off a bit but I was only trying to give a maximum).

-d,d,19 20*6 polynomials with ordering changes like this ( edited to include +0 and -0)
d,-(d-1),18 where d is positive, ....

etc.

edit2: turns out there are 400 possibilities to look through ( as some have only one order that is unique better than searching all 59319 {a,b,c} in that range by hand though.

edit 2 + :
Code:
my(a=[-19..19]);b=setbinop((x,y)->concat(x,y),a);b=setbinop((x,y)->concat(x,y),b,a);b=select(r->vecsum(r)==19,b)

Last fiddled with by science_man_88 on 2017-04-07 at 01:22
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Old 2017-04-07, 02:09   #3
carpetpool
 
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So just looking on the conditions mod 3 we have:

a = 1
b = 2
c = 1

a = 0
b = 0
c = 1

I also thought about another exception: n may also define properties of the cyclotomic field K if and only if each prime power p^k dividing n is either 0 or 1 mod n.

So in this sense, factors such as 2^(2*n), 5^(2*n), 11^(2*n), 17^(2*n), 23^(2*n), 29^(2*n)..., etc. would be allowed, but I don't know weather this would make problem harder.

Depending on (prime) K, we also allow n to divide:

for K = 5, 2^(4*n), 3^(4*n), 7^(4*n), 13^(4*n), 17^(4*n), 19^(2*n), 23^(4*n), 29^(2*n)..., etc.

for K = 7, 2^(3*n), 3^(6*n), 5^(6*n), 11^(3*n), 13^(2*n), 17^(6*n), 19^(6*n), 23^(3*n)..., etc.

for K = 11, 2^(10*n), 3^(5*n), 5^(5*n), 7^(10*n), 13^(10*n), 17^(10*n), 19^(10*n), 29^(10*n)..., etc.

for K = 13, 2^(12*n), 3^(3*n), 5^(4*n), 7^(12*n), 11^(12*n), 17^(6*n), 19^(12*n), 23^(6*n), 29^(4*n)..., etc.

and so on...

Last fiddled with by carpetpool on 2017-04-07 at 02:19
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Old 2017-04-07, 02:15   #4
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Quote:
Originally Posted by carpetpool View Post
So just looking on the conditions mod 3 we have:

a = 1
b = 2
c = 1

a = 0
b = 0
c = 1

I also thought about another exception: n may also define properties of the cyclotomic field K if and only if each prime power p^k dividing n is either 0 or 1 mod n.

So in this sense, factors such as 2^2, 5^2, 11^2, 17^2, 23^2, 29^2..., etc. would be allowed, but I don't know weather this would make problem harder.
well with one more command combination in PARI/gp you'll find that 386 of those 400 are considered irreducible.

Last fiddled with by science_man_88 on 2017-04-07 at 02:16
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