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

  • Thread starter RossH
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  • #1
RossH
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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?
 

Answers and Replies

  • #2
Dick
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9*49=(3*3)*(7*7)=(3*7)*(3*7)=(21)*(21). How did I do that?
 
  • #3
RossH
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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
Dick
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(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.
 

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