A Equation with three consecutive prime numbers

  • Thread starter Dacu
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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.
A solution is [tex]n=2,p_2=3,p_3=5,p_4=7.[/tex]
May be other solutions?


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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])
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]
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