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Math Help - A pretty hard problem .. please help

  1. #1
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    Question A pretty hard problem .. please help

    I have to solve that equation :

    11n-2n=k2

    where n,k are from N .
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  2. #2
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    Re: A pretty hard problem .. please help

    Well by observation n=1, k=3 is one solution. I don't think there are other solutions but I have no way of proving their aren't
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  3. #3
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    Re: A pretty hard problem .. please help

    Yes , I have also find that . And n can be also =0 and k=0 .
    So there are only 2 solutions,I think .. but how should I demonstrate ?
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  4. #4
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    Re: A pretty hard problem .. please help

    0 is generally not considered a natural number
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  5. #5
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    Re: A pretty hard problem .. please help

    Quote Originally Posted by Shakarri View Post
    0 is generally not considered a natural number
    "There is no universal agreement about whether to include zero in the set of natural numbers: some define the natural numbers to be the positive integers {1, 2, 3, ...}, while for others the term designates the non-negative integers {0, 1, 2, 3, ...}. The former definition is the traditional one, with the latter definition having first appeared in the 19th century. Some authors use the term natural number to exclude 0 and whole number to include it; others use whole number in a way that includes both 0 and the negative integers, i.e., as an equivalent of the integer term."

    See this web entry.

    I have a great many different textbooks on set theory. In the vast majority of them use 0\in\mathbb{N}.
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  6. #6
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    Re: A pretty hard problem .. please help

    It's one of the things in mathematics that irks me
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  7. #7
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    Re: A pretty hard problem .. please help

    Quote Originally Posted by Plato View Post
    "There is no universal agreement about whether to include zero in the set of natural numbers: some define the natural numbers to be the positive integers {1, 2, 3, ...}, while for others the term designates the non-negative integers {0, 1, 2, 3, ...}. The former definition is the traditional one, with the latter definition having first appeared in the 19th century. Some authors use the term natural number to exclude 0 and whole number to include it; others use whole number in a way that includes both 0 and the negative integers, i.e., as an equivalent of the integer term."

    See this web entry.

    I have a great many different textbooks on set theory. In the vast majority of them use 0\in\mathbb{N}.
    Ok , firstly , thanks for the explanation . I've read the article about the natural numbers about one year ago
    I haven't seen that the author had mentioned that the numbers are different from 0 . And also , in the some problems that i've seen if the numbers aren't null they are included in N* and in the problem the number are only from N .
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