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Old 2020-11-02, 20:28   #12
bhelmes
 
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Quote:
Originally Posted by bhelmes View Post
tan (alpha)= 11/60,
tan (alpha/2)=(61-60)/11 = 1/11

alpha=arctan (1/11)=5,194428908


sin (alpha)=0,090535746
61²*sin (alpha)=336,883511024
337 =32 mod 61 = 29 mod 61



cos (alpha)=0,995893206
61²*cos (alpha)= 3705,718621265
3705 = 45 = 16 mod 61


29² + 16² = 1 mod 61 (o.k.)


29² - (16i)² = -i²


(29i)² +16² = -1


is this o.k. ?
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Old 2020-11-03, 13:58   #13
Dr Sardonicus
 
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Originally Posted by bhelmes View Post
For mathematical curosity and practical use:

Is it possible to calculate the square root of a quadratic residium by using the tangens function ?
I am unaware of any such method.

For primes p congruent to 1 (mod 4) I don't know of any faster way than factoring x2 - r over the finite field with p elements. In Pari-GP sqrt(Mod(r, p)) will return the square root in (0,p/2) if p is prime and r is a quadratic residue (mod p).
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