 mersenneforum.org > Math calculation of the non quadratic residium
 Register FAQ Search Today's Posts Mark Forums Read  2020-11-02, 20:28   #12
bhelmes

Mar 2016

22×89 Posts Quote:
 Originally Posted by bhelmes 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. ?   2020-11-03, 13:58   #13
Dr Sardonicus

Feb 2017
Nowhere

2×3×5×167 Posts Quote:
 Originally Posted by bhelmes 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).   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post bhelmes Number Theory Discussion Group 3 2020-10-17 13:37 FreakyPotato Programming 7 2015-02-06 10:33 tcharron PrimeNet 4 2014-06-27 23:27 kurtulmehtap Math 3 2010-10-11 15:02 storm5510 Software 8 2009-09-25 21:06

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