abstract algebra – Is ideal $ (x ^ 2 + 2x-3) $ prime in $ mathbb {Z}[X]$?

I have this ideal in $ mathbb {Z}[X]$: $ I = left langle X ^ 2 + 2X-3 right rangle $.

Let's $ P $ and $ Q $ are polynomials: $ P (X) = a_0 + a_1X + a_2X ^ 2 + … $, $ Q (X) = b_0 + b_1X + b_2X ^ 2 + … $.

$ PQ = X ^ 2 + 2X-3 $. Since $ PQ = a_0b_0 + (a_1b_0 + a_0b_1) X + … $, we have the system:

$$ – 3 = a_0b_0 $$

$$ 2 = a_1b_0 + a_0b_1 $$

etc.

I'm not really sure what should be my next step. Can any one help me?

Number Theory – Can the same pattern of recursion define multiple sequences of prime numbers?

First, some definitions:

Let A be a shortcut for an infinite sequence
(a (1), a (2), a (3), …)
just as B is an infinite series of positive integers
(b (1), b (2), b (3), …),
and C is an increasing infinite sequence of positive integers
(c (0), c (1), c (2), c (3), …). Call c (0) the initial value of C.

It is said that the pair A, B is a recursion pattern for C if for each
positive integer n we have
c (n) = a (n) * c (n-1) + b (n). Given a recursion pattern A, B and a
initial value for C, we can reconstruct all the sequence C.

Clearly, for any increasing infinite sequence C, we can find a couple A, B
which is a recursion pattern for C, simply by taking all the a (n) to
be equal to 1 and defining the b as required. If the sequence C is
increasing rapidly, there will be many A, B pairs that will be
recursion models for C.

In particular, for any increasing sequence of prime numbers, one can find
recursion models, many of them if the sequence grows rapidly.

Now the question: can a couple A, B be a recursion pattern for more than
a sequence of prime numbers? In other words, since A and B are a recursion
model, is it possible that there is more than an initial value for c (0)
which will lead to an infinite sequence of prime numbers?

comments
If the answer is yes and we can prove it, I would wait for the proof
to be quite difficult. Maybe conditional proof would be possible.
On the other hand, it can be elementary to show that the answer is no.
This question was written by Moshe Newman.

Number Theory – Prove that if $ p $ is prime and $ a ^ p + b ^ p = c ^ p $, then $ a + b-c = 0 mod p $

I'm working on the little Fermat theorem exercise below:

Prove that if $ p $ is first and $ a ^ p + b ^ p = c ^ p $ then $ a + b-c = 0 mod p $

Also, I find a relation with Fermat's last theorem that says that not three positive integers $ a, b $, and $ c $ satisfy the equation $ a ^ n $ + $ b ^ n $ $ = $ $ c ^ n $ for any integer value of $ n $ bigger than $ 2 $.

So, is there a solution or a way to solve the problem based on the last theorem? How should I go ahead with this exercise? Any suggestion or help will be appreciated.

Find the total number of prime factors in the product {(9) 13 X 5 ^ 4 X (55) ^ 8} a. 73 b. 67 c. 70 d. 63

Find the total number of prime factors in the product {(9) ^ 13 X 5 ^ 4 X (55) ^ 8}
a. 73
b. 67
c. 70
re. 63
spoiler alert the answer is not 46
help would be appreciated

Secret of 1 division to any prime number

Lets see the secrets of prime numbers
when I divide 1 into prime numbers, I see the same occurrence, for example 1/13.
Our number is 0.076923.

76923 x 1 = 076,923

76923 x 2 = 153,846

76923 x 3 = 230,769

76923 x 4 = 307,692

76923 x 5 = 384,615

76923 x 6 = 461,538

76923 x 7 = 538,461

76923 x 8 = 615384

76923 x 9 = 692307

76923 x 10 = 769230

76923 x 11 = 846153

76923×12 = 923076

76923 x 13 = 999999

can you see the iteration of the same numbers 076923
and did you notice the 076 + 923 = 999

So why is it iterates ???

Plus, I did it for other prime numbers small or very big and I got stranger results 🙂 Let's look ..

Secrets of Prime Numbers

The fastest way to find all the prime factors of a very large number without a calculator?

  1. The largest possible prime factor of a very large number would be
    number itself.
  2. The second largest would be n / 2 (if first)
    if n / 2 is not prime, it will be less than n / 2.
  3. The smallest factor would be 1.

Is there a general approach to finding all the remaining prime factors using these three numbers?

Thank you..

Elementary Number Theory – If $ m $ is odd and not prime

Prove that he must have at least one prime factor $ a leq sqrt {m} $ and that all of its prime factors must be at most $ m / $ 3?

Note: I have been able to prove that every non-premium $ m $ has at least one prime divider. Am I correct in assuming that the evidence of the first part will be similar? What I've tried, it's to express $ m $ in terms of two numbers $ a_ {1}, a_ {2} $ and assuming that one of them is greater than $ sqrt {m} $ to reach a contradiction but that seems incomplete.

performance – The next prime number in Python

Have a good day. I've written a piece of code that finds the next prime number after a given number (input), but here I'm looking to optimize it or, more importantly, to run it for n & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; any number up to that 1000000000 in 2 seconds or less.

Here is my code:

of functools import reduce

num = int (input ())

def (n) factors:
returns set (collapse (list .__ add__,
([i, n//i] for i in the range (1, int (n ** 0.5) + 1) if n% i == 0)))

f = (factors (num))

if (len (f)! = 2):
while (len (f)! = 2):
num + = 1
f = factors (num)
print (num)
other:
print (num)

For those who are still wondering, I do it for the problem https://dmoj.ca/problem/bf3 of DMOJ. As a comment, I also found another way to perform exactly the same thing, but this one is slower, so I would not mind worrying about it that way. However, just for reference, I still add the slowest: https://repl.it/@AriAri/Next-Prime-Way-2.

Thank you.

Check if a number is prime

Using the primality test on this site, I found that concatenating the inversion of the figures from the first 548 odd prime numbers in the reverse order is paramount! This is only a prime number from 1998, but it took more than an hour for the site calculator to say it was a prime number.

The calculation was very slow. Could someone here confirm that this result is correct?

The result obtained is 7693749334931393 … 91713111753.

Number Theory – Limits for the number of integers not generated by a subset of prime numbers

For $ p $ a prime number, $ x $ a positive real number greater than $ p $ and k $ an even positive integer, say a maximum subset $ A $ prime numbers not exceeding $ x $containing $ p $ and such as the distance between two consecutive elements of $ A $ Is at least k $ to the $ (k, p, x) $property -gap. Now for such a set $ A $ let $ N_ {A} (x) $ the set of integers not exceeding $ x $ whose elements are products of elements of $ A $. Finally leave $ h_ {A} (x) $ to be the cardinal of $ mathbb {N} cap ([2,x] setminus N_ {A} (x)) $ .

What is the best upper limit for $ inf {h_ {A} (x), a text {a} (k, p, x) text {-gap property} } $ in terms of k $, $ p $ and $ x $?