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Math Help - natural numbers

  1. #1
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    natural numbers

    Hi;

    Find all natural numbers n for which \left[\frac{n^2}{5}\right] is a prime number, where [x] stands for the greatest integer that is less or equal to x.

    Any hint as to where to start appreciated.
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  2. #2
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    If you set \left[\frac{n^2}{5}\right]=p\text{, then }n^2=5p+r where r is 0, 1, 2, 3, or 4.

    Here are a couple hints:

    Some values of r work and some don't.

    If ab=5p, (a>0, b>0, p prime) then either:
    a=5 and b=p
    a=p and b=5
    a=5p and b=1, or
    a=1 and b=5p

    Post again if you still have trouble. It's a really interesting problem.
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  3. #3
    Super Member Bacterius's Avatar
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    Do you mean the floor function ? Like this :

    \left \lfloor \frac{n^2}{5} \right \rfloor = p
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  4. #4
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    Quote Originally Posted by Bacterius View Post
    Do you mean the floor function ? Like this :

    \left \lfloor \frac{n^2}{5} \right \rfloor = p
    Yes, the original poster defined the \left[x\right] notation.
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  5. #5
    Super Member Bacterius's Avatar
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    I'd rather use the real notation instead of defining a new notation for something that already exists. But yeah, he defined it, but I just wanted to make sure. To use the floor function : \left \lfloor [contents] \right \rfloor.
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  6. #6
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    I guess both notations are considered valid, and yes, I meant the floor function.

    Quote Originally Posted by hollywood View Post
    If you set \left[\frac{n^2}{5}\right]=p\text{, then }n^2=5p+r where r is 0, 1, 2, 3, or 4.

    Here are a couple hints:

    Some values of r work and some don't.

    If ab=5p, (a>0, b>0, p prime) then either:
    a=5 and b=p
    a=p and b=5
    a=5p and b=1, or
    a=1 and b=5p

    Post again if you still have trouble. It's a really interesting problem.
    Well thanks, not sure if I fully understand what you wrote, I think it may be better to assume that n=5p+r where r is 0, 1, 2, -1, -2 since the floor function will further let us get rid of the "remainder" part ( \frac{4}{5}<1).
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  7. #7
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    So you would get n^2=25p^2+10pr+r^2, where r = 0, 1, 2, -1, or -2. Then \left\lfloor\frac{n^2}{5}\right\rfloor=5p^2+2pr=p(  5p+2r).

    So this expression must be prime, which means that either p=1 or 5p+2r=1. In the second case, 5p=1-2r and only r=-2 works, but then p=1 and p(5p+2r)=1, which is not a prime.

    For the p=1 case, only three of the possibilities for r work, giving you three possible solutions.
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