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Math Help - Euler's Lucky number

  1. #1
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    Euler's Lucky number

    Euler's Lucky numbers are positive integers n such that m2 − m + n is a prime number for m = 0, , n − 1.

    How do we prove that if n is lucky, then n is prime?
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  2. #2
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    Try proving this by assuming to the contrary that n is composite, or 1.
    Last edited by melese; July 6th 2010 at 05:35 PM. Reason: minor addition
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Interesting post!

    In my(poor) opinion to prove this we can't do what mr. melese offered...

    Here is a link I think it be useful! Conjecture 17. The Ludovicus conjecture about the Euler trinomials
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    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    In my(poor) opinion to prove this we can't do what mr. melese offered..
    If  n wasn't prime, then  n=ab where  n>a>1 .

    Now let  m=a :  a^2-a+ab = a(a-1+b) which is composite. Thus  n isn't lucky. So by contraposition the claim follows.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by chiph588@ View Post
    If  n wasn't prime, then  n=ab where  n>a>1 .

    Now let  m=a :  a^2-a+ab = a(a-1+b) which is composite. Thus  n isn't lucky by contraposition.
    How you can choose m like that?
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    How you can choose m like that?
     m takes on all values less than  n .
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    Quote Originally Posted by chiph588@ View Post
    If  n wasn't prime, then  n=ab where  n>a>1 .

    Now let  m=a :  a^2-a+ab = a(a-1+b) which is composite. Thus  n isn't lucky. So by contraposition the claim follows.
    For completness, when you start with n not a prime you should also examine n=1 (not a prime). Then what happens is that m=0 and m^2-m+n=1, and also n isn't lucky.
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by melese View Post
    For completness, when you start with n not a prime you should also examine n=1 (not a prime). Then what happens is that m=0 and m^2-m+n=1, and also n isn't lucky.
    To avoid this special case you could just assume  n is composite and let  m=0 .
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  9. #9
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    But by definition 1 isn't a composite number (nor a prime), so 1 has a unique place.
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  10. #10
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by melese View Post
    But by definition 1 isn't a composite number (nor a prime), so 1 has a unique place.
    But my way above takes care of 1 just like you did.
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  11. #11
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    I found in a different book that 1 \leq m \leq n-1.
    It said the first 6 lucky numbers are 2, 3, 5, 11, 17 and 41 .
    It should make sense to have m=0 since 0^2-0+2 = 2, which is prime.
    Last edited by novice; July 6th 2010 at 07:17 PM.
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    Quote Originally Posted by chiph588@ View Post
    But my way above takes care of 1 just like you did.
    I assume you mean:
    Quote Originally Posted by chiph588@ View Post
    If  n wasn't prime, then  n=ab where  n>a>1 .

    Now let  m=a :  a^2-a+ab = a(a-1+b) which is composite. Thus  n isn't lucky. So by contraposition the claim follows.
    I don't see how the case n=1 is covered, because n=ab and a>1.
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  13. #13
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by melese View Post
    I assume you mean:

    I don't see how the case n=1 is covered, because n=ab and a>1.
    No I offered another solution later.
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    Quote Originally Posted by chiph588@ View Post
    If  n wasn't prime, then  n=ab where  n>a>1 .

    Now let  m=a :  a^2-a+ab = a(a-1+b) which is composite. Thus  n isn't lucky. So by contraposition the claim follows.
    I spent half a day trying it many different ways. I attempted to prove this by contradiction at the beginning, but to prove m^2-m+n \Rightarrow n being a composite is rather difficult because there isn't a mathematical expression for prime numbers, so I decided to prove it by contrapositive, but my main difficulty was here:

    I let n=ab where 1<a<n and 1<b<n and a,b \in \mathbb{N}.
    Since 1 \leq m \leq n-1, I let m=ab-1.

    I ended up with m^2-m+n= (ab-1)^2-(ab-1)+ab= a^2b^2-2ab+2, which led me to nowhere, and I decided to post the question.

    I found your technique quite interesting. I see why you chose m=a. It makes sense since 1<a<n. That's quite creative.
    Last edited by novice; July 7th 2010 at 06:05 AM.
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  15. #15
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    To show lucky => prime, simply let m = 0

    clearly prime numbers are not necessarily lucky (n=7)

    Are there infinite many lucky number?
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