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Old 2021-07-28, 00:51   #1
bhelmes
 
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Default (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 ?


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Old 2021-07-28, 01:59   #2
Dr Sardonicus
 
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Quote:
Originally Posted by bhelmes View Post
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
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Old 2021-07-29, 20:09   #3
bhelmes
 
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Default (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
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Old 2021-07-31, 13:37   #4
Dr Sardonicus
 
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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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