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A Equation with three consecutive prime numbers

  1. Apr 11, 2016 #1
    Solve the equation [tex]np_n+(n+1)p_{n+1}+(n+2)p_{n+2}=p^2_{n+2}[/tex] where [tex]n\in \mathbb N^*[/tex] and [tex]p_n , p_{n+1} , p_{n+2}[/tex] are three consecutive prime numbers.
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    A solution is [tex]n=2,p_2=3,p_3=5,p_4=7.[/tex]
    May be other solutions?
     
  2. jcsd
  3. Apr 11, 2016 #2

    stevendaryl

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    Well, you know that there can't be a solution with [itex]n \geq 10[/itex]. So there are only 10 possibilities to check.

    Here's my reasoning:
    [itex]n p_n + (n+1) p_{n+1} + (n+2) p_{n+2} \leq 3 (n+2) p_{n+2}[/itex]

    So if [itex]n p_n + (n+1) p_{n+1} + (n+2) p_{n+2} = p_{n+2}^2[/itex], that means

    [itex]3 (n+2) p_{n+2} \geq p_{n+2}^2[/itex]

    which means

    [itex]3 (n+2) \geq p_{n+2}[/itex]

    I'm pretty sure that for [itex]n \geq 10[/itex],
    [itex]3(n+2) \lt p_{n+2}[/itex]

    (When [itex]n=10[/itex], [itex]3(n+2) = 36[/itex] and [itex]p_{n+2} = 37[/itex])
     
  4. Apr 12, 2016 #3
    My reasoning:
    From the original equation we get [tex](n+2)^2+4np_n+4(n+1)p_{n+1}=m^2[/tex] where [tex]m\in \mathbb N^*.[/tex]
    Is there such a natural number [tex]m?[/tex]
     
    Last edited: Apr 12, 2016
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