20170214, 00:54  #1 
"Sam"
Nov 2016
5×67 Posts 
Polynomial Discriminant is n^k for an n1 degree polynomial
Is there a simple way to find an n1 degree polynomial with discriminant D = n^k, (defined with one variable x) and n is prime?
I am not talking about the polynomials of the (simple) form(s) (a^nb^n)/(ab) or (a^n+b^n)/(a+b) as those polynomial forms are excluded from this thread, they can be expanded into n1 degree polynomials already. The property I am getting at is the n1 degree polynomial has factors either 0 or 1 (mod n) when the polynomial is evaluated for any "x" value. Some examples for n = 3 (which don't have the simple form(s) (a^3b^3)/(ab) or (a^3+b^3)/(a+b)) 19x^2+7x+1 37x^2+11x+1 ..... Which come from the discriminant D = b^24ac for ax^2+bx+c. Similarly, WITHOUT searching for coefficients, are there good examples of degree 4, discriminant a power of 5, and of degree 6, discriminant a power of 7. Again, these polynomial must not have any of the Generalized Mersenne form(s) (a^nb^n)/(ab) and (a^n+b^n)/(a+b) The discriminant D is at: For degree 4, integer solutions (a, b, c, d, e) such that: 256 a^3 e^3  192 a^2 b d e^2  128 a^2 c^2 e^2 + 144 a^2 c d^2 e  27 a^2 d^4 + 144 a b^2 c e^2  6 a b^2 d^2 e  80 a b c^2 d e + 18 a b c d^3 + 16 a c^4 e  4 a c^3 d^2  27 b^4 e^2 + 18 b^3 c d e  4 b^3 d^3  4 b^2 c^3 e + b^2 c^2 d^2 = 5^n Computed here. For degree 6, integer solutions (a, b, c, d, e, f, g) such that: Too long for thread, is here. For higher goals, for all primes n < 1000, (which is extremely complicated to do). Like always, hints, suggestions, and answers are appreciated. Last fiddled with by carpetpool on 20170214 at 00:57 
20170214, 01:46  #2  
"Forget I exist"
Jul 2009
Dartmouth NS
2102_{16} Posts 
Quote:
121(4*37) = 27 again not technically a power of three ( it is a power of 3 though). 

20170214, 17:00  #3 
"Forget I exist"
Jul 2009
Dartmouth NS
2102_{16} Posts 
as for the property about 0 and 1 mod n only as factors we have restrictions for example:
all look fine until you realize that x or y in each case has some restriction in number 3 for example, neither x nor y can be of any other remainders except 0 and 1 mod 3 ( mod n respectively). in the second one x has to be of remainder 0 or 1 mod 3 ( mod n respectively), and in the first two cases y can't be odd ( in the first case x can also not be odd). Last fiddled with by science_man_88 on 20170214 at 17:49 
20170215, 17:04  #4 
(loop (#_fork))
Feb 2006
Cambridge, England
2×7×461 Posts 
Quadratics can't have discriminant 3^t for odd t, because b^2 would have to be three mod four, but for even t you can just take b any sufficiently large odd square and factor (3^2t)b^2
E.g. 4x^2+29x+7 has discriminant 729; 2x^2+83x+41 has discriminant 3^8 
20170216, 01:18  #5 
Feb 2017
Nowhere
2·3^{2}·5·71 Posts 
The property I am getting at is the n1 degree polynomial has factors either 0 or 1 (mod n) when the polynomial is evaluated for any "x" value.
What is important here is the number field, not the particular polynomial used to define it. By rescaling the variable, you can also sneak in finitely many primes which violate the property. For example, x^2 + 5*x + 25 has D = 3 * 5^2 so defines the same quadratic field as x^2 + x + 1, but assumes values divisible by 5 (in fact by 25) whenever x is an integer divisible by 5. Similarly, x^2 + 7*x + 49 has the requisite property, but discriminant D = 3 * 7^2, which has a prime factor other than 3. Similarly, the quartic polynomial x^4 + 5*x + 5 has D = 5^3 * 11^2, but defines the same field as the cyclotomic polynomial x^4 + x^3 + x^2 + x + 1, and has the requisite property above. [In this case, I chose the "extraneous" factor 11 to be congruent to 1 (mod 5).] So it is prudent to allow for finitely many exceptional primes in the above property [also for finitely many "extraneous" square factors in the discriminant]. This allows rescaling to produce a monic polynomial [lead coefficient 1] defining the same field as any given (nonzero, nonlinear, irreducible) polynomial in Z[x]. Calling the polynomial f(x), and assuming it to be irreducible, monic and in Z[x], and using p instead of n (since you want n to be a prime), one interpretation of the above property is as follows: With at most finitely many exceptions, if q is a prime not congruent either to 1 or 0 (mod p), then the reduction f(x) (mod q) has no linear factors. For if q is a prime dividing f(r), r a rational integer, then (x  Mod(r, q)) is a linear factor of the reduction of f(x) (mod q). Alternatively (with possibly finitely many exceptions), if f(x) (mod q) has a linear factor, then q is congruent to 1 (mod p). Now the cyclotomic field K of pth roots of unity is characterized by the fact that the primes q which split into p1 distinct prime ideal factors in K/Q are precisely those which are congruent to 1 (mod p). This property extends to the cyclotomic field of mth roots of unity, for any m > 2. The degree over Q is phi(m), which is less than m  1 when m is composite. Fortuitously, cyclotomic polynomials have discriminant equal to the discriminant of the number field they define  no extraneous prime factors. Number fields whose (field) discriminants only have one (finite) prime factor are unusal. BTW, Stickelberger's criterion says that the discriminant of any monic polynomial in Z[x] is congruent either to 1 or 0 (mod 4). Thus, polynomial discriminants equal to an odd power of any prime congruent to 3 (mod 4) are impossible. If p is a prime congruent to 3 (mod 4), the discriminant of the cyclotomic polynomial for the primitive pth roots of unity is D = p^(p  2). 
20170216, 03:46  #6 
"Sam"
Nov 2016
5·67 Posts 
Dr. Sardonicus, Thanks for solving my math problems. Your posts are greatly appreciated around here.
If you know of a degree 6 polynomial which has factors 0 or 1 mod 7, please post examples here. Meanwhile, a degree 4 polynomial x^4+7x^3x^27x+1 with Discriminant D = 415125 = 3^4*5^3*41 is 0, 1 or 4 (mod 5) for any x value and appears to have similar properties to the cyclotomic polynomial other than factors congruent to 4 (mod 5) of x^4+2x^3+4x^2+8x+16, a generalization of x^4+x^3+x^2+x+1 of the form a^4+ab^3+a^2b^2+a^3b+b^4. Would there be an easy way to generate polynomials with number fields the same as the cyclotomic polynomial of degree 4 or more generally n1? EDIT: For x = 32, 1276705 = 5*19*89*151 Last fiddled with by carpetpool on 20170216 at 03:52 
20170216, 15:55  #7 
Feb 2017
Nowhere
2·3^{2}·5·71 Posts 
Would there be an easy way to generate polynomials with number fields the same as the cyclotomic polynomial of degree 4 or more generally n1?
Yes  given the right software [e.g. PariGP], and understanding what it does. Your polynomial, f=x^4+7*x^3x^27*x+1 is irreducible with [according to PariGP] Galois group D4. I also used Pari to find a "reduced" polynomial defining the same field, g = x^4  2*x^3  6*x^2 + 7*x + 11 with D = 5^3 * 41 The polynomial f (mod 3) has a repeated quadratic factor because of the "extraneous" factor 3^4 in the discriminant. The polynomial g (mod 3) splits into 2 distinct quadratic factors. The splitting field contains the quadratic fields defined by t^2  t  1, t^2  t  10, and t^2  t  51 [the last being a defining polynomial for Q(sqrt(D))] Both f and g split into quadratic factors in the field defined by t^2  t  1; g factors as (x^2  x + (t  3)) * (x^2  x + (t  4)) where t^2  t  1 = 0. This means that f and g (mod p) will factor into two quadratic factors whenever 5 is a quadratic residue (mod p), i.e. p == 1 or 1 (mod 5). However, whether either quadratic factor splits further (mod p) is a more complicated proposition; it cannot be described by rational integer congruences (this is because D4 is not an Abelian group). For example, one of the quadratic factors (mod 11) splits into linear factors; both quadratic factors (mod 31) remain irreducible; both quadratic factors (mod 131) split into linear factors. If p is congruent to 2 or 3 (mod 5), g (mod p) either remains irreducible, or splits into irreducible quadratic factors. These cases are not describable with rational integer congruences. I decline to generate more defining polynomials for the field of 7th roots of unity. Last fiddled with by Dr Sardonicus on 20170216 at 16:01 
20170217, 01:10  #8 
"Sam"
Nov 2016
14F_{16} Posts 
Here is an example for quartic polynomials, using Wolfram Alpha I solved for the last three coefficients.
x^4+3x^3+4x^2+2x+1 with discriminant D = 125 has the same number field properties as the cyclotomic polynomial x^4+x^3+x^2+x+1. Is PARI GP capable of doing exactly this for degree 6 polynomials, and finding polynomials with the same number field properties as x^6+x^5+x^4+x^3+x^2+x+1. One way would be setting up a few queries via PARI GP to solve for: Discriminant [ax^6+bx^5+cx^4+dx^3+ex^2+fx+g] for a = 1, 20; b = 1, 20; c = 1, 20; d = 1, 20; e = 1, 20; f = 1, 20; g = 1, 20; where 1, 20 denotes all integers 1 <= (a, b, c, d, e, f, g) <= 20. The number of combinations of polynomials would be, 20^7 = 1280000000 such polynomials searching for discriminant = 16807 or some other related value. Please show me a command via Pari GP which can do five variable solution for: Discriminant of x^6+4x^53x^4+ax^3+bx^2+cx+d = 16807 There should be at least one set of integers which will work with the following property. This is interesting, especially for getting into polynomials of degrees 6, 10, and 12. 
20170217, 01:26  #9  
"Forget I exist"
Jul 2009
Dartmouth NS
10000100000010_{2} Posts 
Quote:
Last fiddled with by science_man_88 on 20170217 at 01:51 

20170217, 02:02  #10 
"Sam"
Nov 2016
5×67 Posts 
Check this out sm88,
(17:57) gp > polsubcyclo(2201, 5) %33 = [x^5  x^4  880*x^3 + 176*x^2 + 179584*x + 26624, x^5 + x^4  28*x^3 + 37*x^2 + 25*x + 1, x^5  x^4  880*x^3 + 6779*x^2 + 14509*x  112039, x^5 + x^4  12*x^3  21*x^2 + x + 5, x^5  x^4  880*x^3 + 15583*x^2  95541*x + 196101, x^5  x^4  880*x^3  2025*x^2 + 49725*x  112039] I figured out the idea to this: polsubcyclo(n, d) will give ALL polynomials with the same number field as the cyclotomic polynomial d if and only if d  phi(n). I've organized the new polynomials. Dr. Sardonicus should take a look at these ones: x^5  x^4  880*x^3 + 176*x^2 + 179584*x + 26624 x^5 + x^4  28*x^3 + 37*x^2 + 25*x + 1 x^5  x^4  880*x^3 + 6779*x^2 + 14509*x  112039, x^5 + x^4  12*x^3  21*x^2 + x + 5 x^5  x^4  880*x^3 + 15583*x^2  95541*x + 196101 x^5  x^4  880*x^3  2025*x^2 + 49725*x  112039 All 5 appear to have the same number field properties as x^5+x^4+x^3+x^2+x+1 EDIT: The mistake is the polynomials are degree 5, not 4! Command: (17:54) gp > ?? polsubcyclo() polsubcyclo(n,d,{v = 'x}): Gives polynomials (in variable v) defining the subAbelian extensions of degree d of the cyclotomic field Q(zeta_n), where d  phi(n). If there is exactly one such extension the output is a polynomial, else it is a vector of polynomials, possibly empty. To get a vector in all cases, use concat([], polsubcyclo(n,d)). The function galoissubcyclo allows to specify exactly which subAbelian extension should be computed. The library syntax is GEN polsubcyclo(long n, long d, long v = 1), where v is a variable number. Last fiddled with by carpetpool on 20170217 at 02:15 
20170217, 02:32  #11 
"Sam"
Nov 2016
5×67 Posts 
One pari command I am searching for is polynomials with the same properties as their cyclotomic polynomials, and finds them based on fixed coefficients in order.
({a, b}, x) would search for a polynomial with the same properties as the cyclotomic polynomial of x, with leading coefficients a, b, c for each higher degree. For example: ({1, 4}, 5) should find all polynomials with the same number field properties as x^4+x^3+x^2+x+1 of the form x^4+4x^3+ax^2+bx+c. That is should output the solutions, if any {a, b, c} for the degrees 2, 1, 0. I've scoured a whole column of polynomial commands and not one serves for this purpose. 
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