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Proposition needed to be proven

  1. Sep 22, 2007 #1
    A relatively lengthy proof I am writing for an assignment leads me to a proposition (which I need to turn into a lemma for the proof to be complete) conjecturing that for any [tex]n \in \mathbb{Z^{*}} [/tex], there are no [tex]a, b, c \in \mathbb{Z^{*}}[/tex] such that [tex]4n + 3 = 5^{a}13^{b}17^{c}[/tex]. I haven't been able to find to tackle the problem, any suggestions?
     
    Last edited: Sep 22, 2007
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  3. Sep 22, 2007 #2

    Gib Z

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    I don't understand you question...do you mean that the sequence defined by [tex]a_n = 4n +3 , n \in \mathbb{Z}[/tex] must generate some primes for n other than 5, 13, 17 and 21?

    The digits 4 and 3 add up to 7, a prime. So a value of n that would keep the digits the same already rules the number out for a heap of divisibility tests for small numbers. Since it only rules out small numbers, try small values of n. n=1, a_1 = 7, prime. n=10, a_10 = 43, prime.

    I dont think thats what you are asking, because I'm sure you would have spotted n=1 straight away.
     
  4. Sep 22, 2007 #3

    AlephZero

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    What do you mean by "4n + 3 must have primes other than 5, 13, 17 and 21"?

    21 isn't prime :confused:

    5, 13, 17, 21 are all of the form 4n+1 not 4n+3 :confused:
     
  5. Sep 22, 2007 #4
    Sorry for stating that 21 is prime, it was 4 am here, easy to say gibberish at this time. That said, I have rectified the original question, so please read it.
     
  6. Sep 22, 2007 #5
    (4n+1)(4k+1) = 16nk+4(n+k)+1 = 4m+1. The form is preserved under multiplication.
     
  7. Sep 22, 2007 #6

    AlephZero

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    OK, I read it, but robert Ihnot got there first...

    5 = 13 = 17 = 1 mod 4, so 5^a 13^b 17^c = 1 mod 4.
     
  8. Sep 22, 2007 #7
    Thanks allot, the proof is complete. :smile:
     
  9. Sep 22, 2007 #8

    mathwonk

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    work mod 4. the lhs is 3 mod 4. what aboutn the rhs? 5=1 mod 4, 13=1 mod 4, and also 17=1 mod 4, so the lhs =3 and the rhs =1.
     
  10. Sep 24, 2007 #9

    uart

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    Sorry to sound ignorant but what does the "*" represent in [tex]\mathbb{Z}^{*}[/tex]
     
    Last edited: Sep 24, 2007
  11. Sep 24, 2007 #10
    Non-negative integers.
     
  12. Sep 24, 2007 #11

    uart

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    "Non-negative". Ok thanks.
     
  13. Sep 24, 2007 #12

    Defennder

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    I'm curious as to why you ask this question, Werg22. I vaguely remember using/reading this lemma, as proven by robert Ihnot here, in a maths book. It was titled "Proofs from the BOOK" or something along those lines. I believe the chapter I found it in was something on the representation of integers as sum of primes or something like that.
     
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