Proving an integer is prime

  1. Sep 12, 2005 #1
    Hey there, I've been having some problems trying to prove this:

    "Let p be an integer other than 0, +/- 1 with this property: Whenever b and c are integers such that p | bc, then p | b or p | c. Prove p is prime. [Hint: If d is a divisor of p, say p = dt, then p | d or p | t. Show that this implies d = +/- p or d = +/- 1.]"

    Proving p *can* be prime isn't too difficult, but proving p *must* be prime has really confused me.

    I've tried going down the path, gcd(p,b) = p if p | b. Which means p = pn + bm, but since b = px, p = pn + pxm = p(n + mx) => n + mx = 1. This doesn't seem to get me anywhere though.

    Then I've tried using b = pn, c = pm and tried various manipulations such as b / n = c / m => bm = cn = pmn but I just don't see how I can get it in the form p = dt.

    Now, if I got it to the form p = dt, I'm not sure how I could prove p must be prime. Who's to say its not prime? I don't know it seems like I'm thinking in circles here. (Sorry for not properly defining n,m and x I just assumed they were integers to save time). Any help would be greatly appreciated. Unfortunately my professor has been out of town for the entire week so I've been unable to seek help from him (its still due though of course :P).
     
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  3. Sep 12, 2005 #2

    HallsofIvy

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    Suppose p is NOT a prime. Then p= mn for some integers m and n. Let b= m, c= n.

    bc= mn is divisible by p but neither b nor c are.
     
  4. Sep 12, 2005 #3

    shmoe

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    Use their hint. If p=dt then you know p|d or p|t, assume it's the former. So p|d and d|p, which is larger in absolute value, d or p?
     
  5. Sep 12, 2005 #4
    "Use their hint. If p=dt then you know p|d or p|t, assume it's the former. So p|d and d|p, which is larger in absolute value, d or p?"

    Ok so since p = dt (or nm) p / d = t, since t is an int d | p. If p | d, d / p = 1 / t, so t = +/- 1 and thus d = +/- p. Conversely if p | t, WLOG t = +/- p, d = +/- 1. So d = +/- p or +/- 1 and p is prime. Is this correct?
     
  6. Sep 12, 2005 #5

    shmoe

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    Looks good, though not exactly what I had in mind. If a|b then a<=b, etc (assuming a, b positive-it's such a pain to work with the integers instead of the naturals here, note there should also be a requirement that p is positive if you mean it to be a prime in the usual sense).
     
  7. Sep 12, 2005 #6
    Thanks for the input you two, you've truly helped out.
     
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