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Old 2022-04-10, 06:09   #1
Dobri
 
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May 2018

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Default Factorization of (x^p-1)/(x-1) over the integers modulo n

Let's start this thread with the polynomial (x^p-1)/(x-1) for p = 11, so that
(x11-1)/(x-1) = x10+x9+x8+x7+x6+x5+x4+x3+x2+x+1.
The corresponding composite Mersenne number is M11 = 211-1 = 2047 = 23 × 89.
A complete polynomial factorization is achieved over the integers modulo n = 23 and 89 as follows:
x10+x9+x8+x7+x6+x5+x4+x3+x2+x+1 = (5+x)(7+x)(10+x)(11+x)(14+x)(15+x)(17+x)(19+x)(20+x)(21+x) over the integers modulo 23, and
x10+x9+x8+x7+x6+x5+x4+x3+x2+x+1 = (11+x)(22+x)(25+x)(44+x)(50+x)(57+x)(73+x)(81+x)(85+x)(87+x) over the integers modulo 89.
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Old 2022-04-10, 06:24   #2
sweety439
 
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Can polynomial still be factored over the integers modulo n when n is composite?
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Old 2022-04-10, 06:29   #3
Dobri
 
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Quote:
Originally Posted by sweety439 View Post
Can polynomial still be factored over the integers modulo n when n is composite?
Concerning the Galois fields GF(n), n has to be either a prime number or a prime power.
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Old 2022-04-10, 06:32   #4
sweety439
 
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So if n is neither prime nor prime power, then polynomials cannot be factored mod n?
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Old 2022-04-10, 06:53   #5
Dobri
 
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Quote:
Originally Posted by sweety439 View Post
So if n is neither prime nor prime power, then polynomials cannot be factored mod n?
It can be done, here is an example for p = 16:
(x16-1)/(x-1) = (1+x) (1+x2) (1+x4) (1+x8).
However, note that p = 16 is a composite number.

Last fiddled with by Dobri on 2022-04-10 at 07:13
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Old 2022-04-10, 11:50   #6
Dobri
 
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Let p = 13, then the corresponding Mersenne prime is M13 = 213-1 = 8191.
The polynomial x12+x11+...+x+1 can be fully factored over the integers modulo the prime numbers n = 2kp+1 as follows:
x12+x11+...+x+1 = (x+4)(x+6)(x+7)(x+9)(x+11)(x+17)(x+25)(x+29)(x+37)(x+38)(x+40)(x+43) over the integers modulo the prime number 53 = 2×2×13+1;
x12+x11+...+x+1 = (x+12)(x+14)(x+15)(x+17)(x+27)(x+33)(x+41)(x+57)(x+58)(x+61)(x+69)(x+71) over the integers modulo the prime number 79 = 2×3×13+1;
etc.
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Old 2022-04-10, 12:06   #7
Dobri
 
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Addendum to post #1 (p = 11) for completeness:
23 = 2×11+1 and 89 = 2×4×11+1
because the factors of composite Mersenne numbers are prime numbers n = 2kp+1.
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Old 2022-04-10, 12:16   #8
Dr Sardonicus
 
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Sure, you can factor polynomials modulo n when n is composite. But factorization is not unique.

For example:

Modulo 4 we have x^2 = (x - 2)^2.

Modulo 6 we have x^2 - x = x*(x - 1) = (x + 2)*(x + 3).

Modulo 8 we have x^2 - 1 = (x - 1)*(x + 1) = (x - 3)*(x + 3) = (x - 5)*(x + 5) = (x - 7)*(x + 7).

How bad things can get is left as an exercise for the reader.

EDIT: It is well known that for n > 1, and q a prime number not dividing n, the cyclotomic polynomial \Phi_{n}(x) factors in Fq[x] into polynomials of degree f, where

f is the least positive integer for which qf == 1 (mod n).

Last fiddled with by Dr Sardonicus on 2022-04-10 at 13:15 Reason: as indicated
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Old 2022-04-10, 13:41   #9
Dobri
 
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Let n = p, then xp-1+...+1 = (x+p-1)p-1 over the integers modulo the prime number p.
For example:
x2-1+1 = (x+2-1)2-1 = x+1 over the integers modulo 2;
x3-1+x+1 = (x+3-1)3-1 = (x+2)2 over the integers modulo 3;
...
x11-1+x9+...+x+1 = (x+11-1)11-1 = (x+10)10 over the integers modulo 11;
x13-1+x11+...+x+1 = (x+13-1)13-1 = (x+12)12 over the integers modulo 13;
...
x1277-1+x1275+...+x+1 = (x+1277-1)1277-1 = (x+1276)1276 over the integers modulo 1277;
...
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Old 2022-04-10, 14:28   #10
Dobri
 
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Let p=1277, then the first integer k for which 2kp+1 is a prime is k=10 so that 2×10×1277+1 = 25541.
Thus x1277-1+x1275+...+x+1 can be fully factored over the integers modulo 25541 as shown in the code section below.
Code:
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Old 2022-04-18, 16:17   #11
Dobri
 
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May 2018

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If xp-1+xp-2+...+x+1 = (x+p)(x+...)... over the integers modulo 2kp+1,
then 2kp+1 is a factor of the composite Mersenne number Mp.

Also, if 2kp+1 is a factor of a composite Mersenne number Mp,
then xp-1+xp-2+...+x+1 = (x+p)(x+...)... over the integers modulo 2kp+1.
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∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
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