Graduate Equation with three consecutive prime numbers

[Dacu]

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

 

[stevendaryl]

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

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

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

3 (n+2) p_{n+2} \geq p_{n+2}^2

which means

3 (n+2) \geq p_{n+2}

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

(When n=10, 3(n+2) = 36 and p_{n+2} = 37)

 

[Dacu]

My reasoning:
From the original equation we get (n+2)^2+4np_n+4(n+1)p_{n+1}=m^2 where m\in \mathbb N^*.
Is there such a natural number m?