 mersenneforum.org Phi mod 11 is congruent 4 and 8
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 Register FAQ Search Today's Posts Mark Forums Read  2022-06-25, 20:55   #23
x13420x

Jun 2022

2·5 Posts Quote:
 Originally Posted by retina √2 ≡ 3i ≡ 8i (mod 11)
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....  2022-06-25, 21:01 #24 retina Undefined   "The unspeakable one" Jun 2006 My evil lair 11001110001002 Posts (3i)² = 3² × i² = 9 × -1 = -9 ≡ 2 (mod 11) Last fiddled with by retina on 2022-06-25 at 21:02  2022-06-26, 18:17 #25 retina Undefined   "The unspeakable one" Jun 2006 My evil lair 22·17·97 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?  2022-06-26, 19:10   #26
charybdis

Apr 2020

32×5×19 Posts Quote:
 Originally Posted by retina 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?
The theorem that says a polynomial has at most as many roots as its degree only holds over integral domains, i.e. rings in which it is impossible for two non-zero elements to multiply to zero. Examples of integral domains include the integers, the reals, the complex numbers, the integers mod p for any prime p, the Gaussian integers mod 11... but not the Gaussian integers mod 17.

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)(4-i). So in the Gaussians mod 17, (4+i)(4-i) ≡ 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 4-i). 4+i and 4-i 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)(a-bi), and both a+bi and a-bi have norm p and are therefore Gaussian primes. So a polynomial equation mod p reduces to equations mod a+bi and a-bi, 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.  2022-06-27, 02:20   #27
Dr Sardonicus

Feb 2017
Nowhere

597610 Posts Quote:
 Originally Posted by retina 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?
If by "mod 17" you mean, "an element of the residue ring of the integers Z mod 17Z," the terms "real" and "imaginary" are meaningless. I note that, mod 17, 4^2 +1 == 13^2 + 1 == 0 so 4 and 13 are square roots of -1 (mod 17).

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 2022-06-27 at 02:20 Reason: xinfig topsy  2022-06-27, 09:36 #28 sweety439   "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 72·73 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.  2022-06-27, 09:40   #29
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

72×73 Posts Quote:
 Originally Posted by x13420x Did you guys notice when you add the 2 results you get the modulo number plus 1? like 8 + 4 = 11 + 1
since they are two roots of x^2-x-1, their sum is 1 and their product is -1 (in Z/nZ)

you can see all modulos < 1000

however, for mod 5, there is only one solution (i.e. x^2-x-1 has only root in Z/5Z), and for mod 209, there are four solutions (i.e. x^2-x-1 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  2022-07-02, 20:03 #30 x13420x   Jun 2022 2×5 Posts the sum of all fibonacci numbers is congruent 6 mod 11  2022-07-02, 20:51 #31 Batalov   "Serge" Mar 2008 Phi(4,2^7658614+1)/2 19×523 Posts  Moderator warning to x13420x - stop posting nonsense.  2022-07-02, 21:57   #32
x13420x

Jun 2022

2·5 Posts Quote:
 Originally Posted by Batalov  Moderator warning to x13420x - stop posting nonsense.
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.  2022-07-02, 22:12   #33
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

19×523 Posts Quote:
 Originally Posted by x13420x 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.
Ok, apparently you prefer the second way - a ban for a week to go and in solitude which is a sister of wisdom, ponder on this matra - Ex falso quodlibet.

What you wrote is "if <insert false statement>, why cant an infinite series be the same?"  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post enzocreti enzocreti 1 2020-03-03 06:12 enzocreti enzocreti 1 2020-02-14 16:56 enzocreti enzocreti 2 2020-02-12 15:14 enzocreti enzocreti 0 2020-01-09 11:56 smslca Math 2 2012-01-29 11:30

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