20201031, 03:35  #1 
Mar 2016
2^{4}×3^{3} Posts 
calculation of the non quadratic residium
A peaceful and pleasant night for you,
I know that from the pyth. trippel (3, 4, 5)  > 3/5, 4/5 mod 61 = (13+25i) 1. and that (13+25i)^30 = 1 mod 61 2. and 16+29i = (16²+29²) = 1 mod 61 which is a non quadratic residium How do I calculate 2. from 1. ? Thanks in advance if someone knows a good answer or a link. Last fiddled with by bhelmes on 20201031 at 03:52 
20201031, 11:48  #2 
Dec 2012
The Netherlands
2·7·131 Posts 

20201031, 18:57  #3  
Mar 2016
2^{4}×3^{3} Posts 
Quote:
(16+29i)²=(25+13i) mod 61 (25+13i) can be mirrored at the main diagonale, so that the point of the unit circle (13+25i) "=" (25+13i) tan (alpha)=3/4, tan (alpha/2)=(54)/3=1/3 if this helps I want to calculate the non quadratic residium of (25+13i) mod 61, but i do not know how. Last fiddled with by bhelmes on 20201031 at 18:58 

20201101, 02:58  #4 
Feb 2017
Nowhere
2·31·103 Posts 
PariGP begs to differ:
Code:
? (Mod(13,61)+Mod(25,61)*I)^30 %1 = Mod(60, 61) (Mod(25,61)+Mod(13,61)*I)^30 = Mod(1,61). Your question is complicated by the fact that, in the ring of Gaussian integers R = Z + Z*I, I^{2} = 1 (PariGP notation), 61 is not prime; 61R is not a prime ideal. We have 61R = P_{1}P_{2}, where P_{1} = (5 + 6*I)R and P_{2} = (5  6*I)R are prime ideals. The residue rings R/P_{1} and R/P_{2} are "different" copies of the finite field F_{61} = Z/61Z, the integers modulo 61. A quadratic residue (mod 61R) is, presumably, a quadratic residue (mod P_{1}) and (mod P_{2}). R/P_{1} may be described by mapping a + b*I to a  11*b (a, b integers) and reducing this integer mod 61; for R/P_{2} map a + b*I to a + 11*b and reduce mod 61. Quadratic residues (mod 61R) are characterized by both integer modulo results being quadratic residues (mod 61). Getting from the two ordinary quadratic residues (mod 61) to a square root of your complex number (mod 61R) is a little involved, but can be done. Doing the two mappings with 13 + 25*I gives Mod(43, 61) and Mod(44, 61), both quadratic nonresidues. So 13 + 25*I is a quadratic nonresidue (mod 61R). However, with 25 + 13*I we get Mod(4, 61) and Mod(46, 61) which are both quadratic residues, so 25 + 13*I is a quadratic residue (mod 61R). And in fact, (Mod(16, 61) + Mod(29,61)*I)^{2} = Mod(25,61) + Mod(13,61)*I. 
20201101, 09:25  #5  
Mar 2016
660_{8} Posts 
Quote:
??? I think 61 is a prime ideal in the Gaussian integers, 61=(6+5i)(65i) http://devalco.de/poly_xx+yy_demo.ph...60&radius_c=61 I understand that i²=1 I appreciate your explications and patience. Last fiddled with by bhelmes on 20201101 at 09:26 

20201101, 09:51  #6  
Dec 2012
The Netherlands
2·7·131 Posts 
Quote:
Since 61=(6+5i)(65i), 61 is not irreducible in the Gaussian integers and therefore not prime, hence it does not generate a prime ideal. 

20201101, 15:52  #7  
Feb 2017
Nowhere
2·31·103 Posts 
Quote:
(a) closed under addition (if x and y are in M then so is x + y), and (b) closed under multiplication by all elements of the ring R; if r is any element of R, and x is in M, then r*x is also in M. A prime ideal P in a commutative ring R is an ideal such that, if x and y are in R and x*y is in P, then x is in P or y is in P. With R = Z + Z*i, neither x = 6 + 5*i nor y = 6  5*i is in 61R, but their product is. Therefore, by definition, 61R is not a prime ideal of R. 

20201101, 20:00  #8 
Mar 2016
432_{10} Posts 

20201101, 21:22  #9  
Feb 2017
Nowhere
1100011110010_{2} Posts 
Quote:
The fact that 61R = P_{1}P_{2} has consequences. Typically^{*}, a square Mod(a,61) + Mod(b,61)*I has two square roots (mod P_{1}) and two square roots (mod P_{2}). Consequently, it will typically have four square roots mod 61R. For example, Mod(25,61) + Mod(13,61)*I has the square roots Mod(16, 61) + Mod(29, 61)*I, Mod(14, 61) + Mod(7, 61)*I, Mod(45, 61) + Mod(32, 61)*I, and Mod(47, 61) + Mod(54, 61)*I. ^{*}The exceptional cases are left as an exercise for the reader. 

20201102, 18:40  #10 
Mar 2016
2^{4}×3^{3} Posts 
For mathematical curosity and practical use:
Is it possible to calculate the square root of a quadratic residium by using the tangens function ? I have the pythagoraic tripple 11,60,61 (which indicates a rational point on the circle, tan (alpha)= 11/60, tan (alpha/2)=(6160)/11 = 1/11 and 1=i² 1/11 = (1+9*61) / 11 = 55 mod 61 But then I do not know ? Best regards Bernhard 
20201102, 19:04  #11  
Oct 2020
1_{8} Posts 
Quote:


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