20210728, 00:51  #1 
Mar 2016
2·3·59 Posts 
(p  ax²+bx+c) && (p  Ax²+Bx+C)
A peaceful night for you,
Let f(x)=ax²+bx+c and g(x)=Ax²+Bx+C with discriminant (f(x))=/=discr (g(x)) and f(x0)=g(x0) I am looking for a prime p>2 where p  f(x1) and p  g(x1) Is there a better way to calculate p than calcualting gcd [f(x), g(x)] by increasing x ? 
20210728, 01:59  #2  
Feb 2017
Nowhere
4849_{10} Posts 
Quote:
EDIT: I am assuming that gcd(a, b, c) = gcd(A, B, C) = 1. Last fiddled with by Dr Sardonicus on 20210728 at 12:30 Reason: As indicated 

20210729, 20:09  #3 
Mar 2016
542_{8} Posts 
(p = a(x1)²+b(x1)+c) && (q = A(x1)²+B(x1)+C)
@Dr Sardonicus
It is always illuminating and friendly getting a response from you, thanks a lot for your clear and precious words. If p(x1)=a(x1)²+b(x1)+c and If q(x1)=A(x1)²+B(x1)+C with discr (p(x))=  discr (q(x)) Normally I would make a probablistic prime test for p(x1) and q(x1) seperately in order to check primality for both. Can I combine and speed up the prime checking, assuming that both are primes ? Last fiddled with by bhelmes on 20210729 at 20:12 
20210731, 13:37  #4 
Feb 2017
Nowhere
11361_{8} Posts 
It would help if you kept your notation consistent. You called your quadratic polynomials f(x) and g(x) in one post, then p(x) and q(x) in the next.
I'll stick with f(x) and g(x), and say p = f(x1), q = g(x1). I note that (assuming the coefficients are integers), for two quadratic polynomials to have equal and opposite discriminants, the xcoefficients b and B have to be even. For if b^2  4*a*c + B^2  4*A*C = 0, we have b^2 + B^2 = 4*(a*c + A*C). The only way for the sums of the squares of two integers to be divisible by 4 is, for both squares to be even, so their roots are also even. This does allow for some reformulation, e.g. a*f(x) = (a*x + b/2)^2 + a*c  b^2/4 and A*g(x) = (A*x + B/2)^2 + A*C  B^2/4 with a*c  b^2/4 and A*C  B^2/4 are equal and opposite. I do not know of any way to combine the tests of p = f(x1) and q = g(x1). 