# pr.probability – To prove a relation involving a probability distribution

I’m reading a book and have encountered a relation which seems to me to be impossible to prove, I would like to be sure if this is the case. The author gives a probability function as
$$p_n = frac{e^{-c_1 n – c_2/n}}{Z},$$
where $$c_1$$ and $$c_2$$ are constants and Z is a normalization factor and $$n geq 3$$. Then by defining $$alpha$$ as $$alpha = sum_{n = 3}^{infty} p_n (n – 6)^2$$, the author claims one can show that

$$begin{equation} alpha + p_6 = 1, quad quad quad 0.66 < p_6 < 1, end{equation}$$
$$begin{equation} alpha p_6^2 = 1 / 2 pi, quad quad quad 0.34 < p_6 < 0.66. end{equation}$$

How is such a thing possible in the first place as these relations are not even dependent on $$c_1$$ and $$c_2$$?