$\frac{b^{2n}+b^{n+1}+3b-5}{b-1}$ is square

Question

Find all $b>5$ so that $x_n = \frac{b^{2n}+b^{n+1}+3b-5}{b-1}$ is square for all sufficiently so large integers n. I think the only value of $b$ is 10.

If there is $p \in \mathbb{P}$ (prime), $p \neq 3, p \mid b-1$, then $v_p(b-1)=1$. If $v_p(b-1) > 1$, we can choose n.

If $b=3^m+1$, then m must be even.(If m is odd, $r_8(x_n)=3$)

What should I do next?

Answer 1

COMMENT: May be using these ideas help:

1-Suppose $\frac{b^{2n}+b^{n+1}+3b-5}{b-1}= (b+a)^2$

If polynomial $f(b)=b^{2n}+b^{n+1}+3b-5$ must have a double root like $-a$ then we must have:

$f(-a)=f'(-a)=0$

This gives following system of equations:

$\begin{cases}\frac{(-a)^{2n}+(-a)^{n+1}-3a-5}{-a-1}=0\\2n(-a)^{2n-1}+(n+1)(-a)^n+3=0 \end {cases}$

Which gives a and n.Plugging these in fraction gives an equation in terms of b which gives b.

2- We must have:

$\frac{b^{2n}+b^{n+1}+3b-5}{b-1}=(b+a)^2$

Which gives:

$b^{2n}+b^{n+1}+3b-5=(b-1)(b+a)^2$

Or:

$b^{2n}+b^{n+1}-b^3-(2a-1)b^2-(a^2-2a)b+a^2-5$

Which is a Diophantine equation to be solved.