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 2021-07-28, 00:51 #1 bhelmes     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 ?
2021-07-28, 01:59   #2
Dr Sardonicus

Feb 2017
Nowhere

484910 Posts

Quote:
 Originally Posted by bhelmes 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 ?
Any such p has to divide the resultant of f(x) and g(x). If no prime factor of the resultant provides a solution, there isn't one.

EDIT: I am assuming that gcd(a, b, c) = gcd(A, B, C) = 1.

Last fiddled with by Dr Sardonicus on 2021-07-28 at 12:30 Reason: As indicated

 2021-07-29, 20:09 #3 bhelmes     Mar 2016 5428 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 2021-07-29 at 20:12
 2021-07-31, 13:37 #4 Dr Sardonicus     Feb 2017 Nowhere 113618 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 x-coefficients 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).

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