[number theory] find number in certain domain with two prime factorizations

In summary, if you are trying to find a number in the domain with multiple prime factorizations, you can try multiplying together numbers that take the form 4k-1.
  • #1
RossH
75
0

Homework Statement


My domain i numbers of form 4k+1. n divides m is this domain if n=mk for some k in the domain. A number is prime in this domain if its only divisors are 1 and itself. My problem is to find a number in the domain with multiple prime factorizations.

Homework Equations


None.

The Attempt at a Solution


If have started by identifying ja series of primes in the domain: 5,9,13,17,21,29, and all other numbers that are prime in our domain and are in this domain (I'll call it M-space). So far I have simply been trying to multiply together randomly and hope to get the same number. I have been unsuccessful. Is there any systematic approach that I could try?
 
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  • #2
9*49=(3*3)*(7*7)=(3*7)*(3*7)=(21)*(21). How did I do that?
 
  • #3
Dick said:
9*49=(3*3)*(7*7)=(3*7)*(3*7)=(21)*(21). How did I do that?

(4n-1)(4m-1)=16mn+4n+4m+1=4(4mn+n+m)+1=4k+1
Therefore, any two numbers that take the form 4k-1 will be in the domain when multiplied together, so you only need to find factors that are of this form and rearrange them (as long as they can be factored in the normal domain). For example,

(21*209)=(3*7)(11*19)=(3*11)(7*19)=33*133=4389

Is my logic right? Does this work in the general sense? Thank you for your help. Number theory is driving me out of my mind.
 
  • #4
RossH said:
(4n-1)(4m-1)=16mn+4n+4m+1=4(4mn+n+m)+1=4k+1
Therefore, any two numbers that take the form 4k-1 will be in the domain when multiplied together, so you only need to find factors that are of this form and rearrange them (as long as they can be factored in the normal domain). For example,

(21*209)=(3*7)(11*19)=(3*11)(7*19)=33*133=4389

Is my logic right? Does this work in the general sense? Thank you for your help. Number theory is driving me out of my mind.

Sure. That's exactly it. Number theory was tough for me too. Seemed like just a huge bag of tricks. Sometimes fun though.
 

1. How do you find a number in a certain domain with two prime factorizations?

In order to find a number in a certain domain with two prime factorizations, you can use the fundamental theorem of arithmetic. This theorem states that every positive integer can be written as a unique product of prime numbers. By factoring the given number into its prime factors, you can then compare the two factorizations to find the common primes and determine the number in the given domain.

2. What is the significance of finding a number with two prime factorizations?

Finding a number with two prime factorizations can have significant implications in cryptography and number theory. It can also help in identifying patterns and relationships between numbers, as well as providing a deeper understanding of prime numbers and their properties.

3. Can a number have more than two prime factorizations?

No, according to the fundamental theorem of arithmetic, every positive integer can only have one unique prime factorization. This is because prime numbers are the building blocks of all other numbers, and any composite number can be broken down into a unique combination of prime factors.

4. How can this concept be applied in real-life situations?

The concept of finding a number with two prime factorizations can be applied in various areas such as cryptography, data encryption, and number theory. It can also be used in everyday life to solve mathematical problems and puzzles, as well as in identifying patterns in numbers and sequences.

5. Is there a specific method for finding a number with two prime factorizations?

Yes, there are several methods for finding a number with two prime factorizations. Some common methods include using prime factorization algorithms, factoring by grouping, and the Euclidean algorithm. The most efficient method may vary depending on the size and complexity of the given number.

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