20220625, 20:55  #23 
Jun 2022
2·5 Posts 
Look I have made a lot of mistakes so this could be just another goof up.....but when I convert the infite series of square root of 2 to a fraction I got a division by zero.....like I said before it envolves fractions with factorials in the denomenator and numerator....

20220625, 21:01  #24 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
1100111000100_{2} Posts 
(3i)² = 3² × i² = 9 × 1 = 9 ≡ 2 (mod 11)
Last fiddled with by retina on 20220625 at 21:02 
20220626, 18:17  #25 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·17·97 Posts 
There are 4 square roots mod 17; 2 real and 2 imaginary.
6² ≡ 2 (mod 17) 11² ≡ 2 (mod 17) (7i)² ≡ 2 (mod 17) (10i)² ≡ 2 (mod 17) My maths teacher taught me that square roots only have 2 solutions. They were wrong. Are there any prime moduli that have more than 4 square roots? 
20220626, 19:10  #26  
Apr 2020
3^{2}×5×19 Posts 
Quote:
The Gaussian integers mod 17 are not an integral domain for the same reason that the ordinary integers mod a composite number are not an integral domain. Once you allow i, 17 isn't a prime anymore, as it is equal to 4^2 + 1^2 = (4+i)(4i). So in the Gaussians mod 17, (4+i)(4i) ≡ 0. This will happen for every prime of the form 1 mod 4, as they can all be represented as a sum of two squares. Now the equation x^2 ≡ 2 (mod 17) reduces to the two equations x^2 ≡ 2 (mod 4+i) and x^2 ≡ 2 (mod 4i). 4+i and 4i are both Gaussian primes (they have to be, since their norm 17 is a prime) and so each equation has at most two solutions. In this case both actually do have two solutions, and this results in four solutions mod 17. There will never be more than 4 solutions for a prime modulus. Prime factorization over the Gaussians is unique. If p is 3 mod 4 then p is prime over the Gaussians too, so the Gaussians mod p are an integral domain and a quadratic has at most 2 solutions. If p is 1 mod 4, then p = a^2 + b^2 = (a+bi)(abi), and both a+bi and abi have norm p and are therefore Gaussian primes. So a polynomial equation mod p reduces to equations mod a+bi and abi, and each has at most as many solutions as the degree of the polynomial. Combining these, we get that the number of solutions mod p is at most degree^2. 

20220627, 02:20  #27  
Feb 2017
Nowhere
5976_{10} Posts 
Quote:
And 7*4 == 11 (mod 17), 10*4 == 6 (mod 17), 7*13== 6 (mod 17), 10*13 == 11 (mod 17). OTOH, if you mean by "mod 17" "an element in the residue ring Z[i]/17Z[i]," you're changing the rules. In Z[i], 17 isn't prime, and the residue ring isn't a field (it's the direct product of two copies of Z/17Z). The rules for polynomials with coefficients in a field don't apply. The short answer to your question is "no," assuming you're working in Z[i] or the ring of integers in any quadratic field. If K is a number field, R is the ring of integers in K, K/Q has degree greater than 2, and 17R splits into distinct ideals of degree 1, then the answer in R/17R is "yes." Last fiddled with by Dr Sardonicus on 20220627 at 02:20 Reason: xinfig topsy 

20220627, 09:36  #28 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
7^{2}·73 Posts 
Zp(i) is an quadratic extension field of the field Zp if and only if p == 3 mod 4, if p == 1 mod 4, Zp(i) has zero divisors.

20220627, 09:40  #29  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
7^{2}×73 Posts 
Quote:
you can see all modulos < 1000 however, for mod 5, there is only one solution (i.e. x^2x1 has only root in Z/5Z), and for mod 209, there are four solutions (i.e. x^2x1 has four roots in Z/209Z), for all n < 209 (except n = 1 and n = 5 there is only one solution mod n), there are either two solutions (with sum 1 and product 1) or no solutions 

20220702, 20:03  #30 
Jun 2022
2×5 Posts 
the sum of all fibonacci numbers is congruent 6 mod 11

20220702, 20:51  #31 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
19×523 Posts 
Moderator warning to x13420x  stop posting nonsense. 
20220702, 21:57  #32 
Jun 2022
2·5 Posts 
Why would it be nonsense? If an irrational number that goes to infinity can have a residue mod 11 why cant an infinite series be the same? I really dont understand the diference.....does everyone just assume I am stupid because I made some careless mistakes? E, pi and square root of two do not have a residues mod 11 because of division by 11.......but fibonacci has factors of 11 in the numerator and denomenator so they cancel.

20220702, 22:12  #33  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
19×523 Posts 
Quote:
What you wrote is "if <insert false statement>, why cant an infinite series be the same?" 

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