20200321, 03:25  #23 
Aug 2019
B_{16} Posts 
1. Is the polynomial constant a prime number?
No 2. Is the lead coefficient a prime? [or monic]? monic 3.If not, are these coefficients special in some way? Don't understand this question Please explain why you need/want to do this. 4. tell us what you want to do if there are no rational solutions.  If the equation has no rational solution, we don't need the solution. 
20200321, 03:30  #24  
Aug 2019
11 Posts 
Quote:
It comes from elliptic curve, they are either integers or fraction, there are 4 coefficients which has huge number, if you save them, each of them will take 1 MB in a text file, it is that simple, the polynomial is monic, thanks. 

20200321, 04:16  #25 
Nov 2003
2^{2}×5×373 Posts 

20200321, 12:55  #26  
Feb 2017
Nowhere
2^{4}·3^{2}·29 Posts 
Quote:
If you have any a priori knowledge about how many pairs of complexconjugate roots the equation has, use it. For example, if you know the equation always has only real roots, just take the real part of what polroots() gives you. Another option is to try to reduce the polynomial (mod p) for small primes p which do not divide any of the denominators. If a mod p factorization does not have any linear factors, neither does the given polynomial, and you're done. Each such mod p reduction and factorization can be done fairly quickly, but, alas I can offer no guarantee this method will work for any really small primes p. It also is of no help in finding a rational root if one exists. 

20200321, 15:32  #27  
Aug 2006
3·1,987 Posts 
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20200321, 16:34  #28  
Feb 2017
Nowhere
2^{4}·3^{2}·29 Posts 
Quote:
The method R. Gerbicz outlined for a poly with integer coefficients only requires computing real roots r to within 1/(2*a(n)) where a(n) is the lead coefficient, and then checking whether round(r*a(n))/a(n) is a root. That's still potentially a lot of real precision, along with the large coefficients, but that should be no problem. Since we're only dealing with quartics, this approach seems quite practicable. 

20200321, 17:01  #29  
Nov 2003
2^{2}×5×373 Posts 
Quote:
that I don't understand. The OP seems to know at least some math since he/she is playing with elliptic curves. However, it puzzles me why he/she would ask for rational roots when he/she would know that the root is an integer........ It would be nice to know where the problem comes from. 

20200321, 17:19  #30  
Feb 2017
Nowhere
2^{4}×3^{2}×29 Posts 
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20200321, 17:23  #31 
Nov 2003
2^{2}×5×373 Posts 

20200321, 18:07  #32 
Feb 2017
Nowhere
2^{4}·3^{2}·29 Posts 
Indeed. With my very limited knowledge of elliptic curves, only one possibility occurs to me: If the quartic f(x) has a rational zero r, then (0, r) is a readymade rational point on the curve
y^{2} = f(x) which can be used to transform the curve into an elliptic curve y^2 = F(x), F cubic. If f(x) has no rational zero, there is no rational point (0, r) on the curve. Of course if f(x) has no rational zeroes, it is still possible AFAIK that y^{2} = f(x) could have some rational point with y <> 0. EDIT: Of course, with y <> 0 , there would be two rational points (y, r) and (y, r) whereas with y = 0 you only get one ratonal point. Last fiddled with by Dr Sardonicus on 20200321 at 18:13 Reason: As indicated 
20200321, 19:36  #33 
Aug 2019
11 Posts 

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