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.
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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 $$m?$$