Solutions to the Diophantine equation

Question

This cropped up in an otherwise simple-looking problem. Find the solutions for $a, b, n \in \mathbb{Z}$ and $b, n > 1$ for the Diophantine equation:

$b^n + 1 = a^2$

Alternatively:

$a^2 - b^n = 1$

One can see that if $n$ is even there are no solutions. But for $n$ odd, there can be solutions, one of which is of course evident in $3^2 - 2^3 = 1$

Is this an open problem or do we know the solutions here?

Edit: Is there a simple, elementary solution to this special case?

Answer 1

HINT.- It seems that the only solution is $(a,b,n)=(2,3,1)$. In fact $b^n=(a+1)(a-1)$ so you can do $a+1=r^n$ and $a-1=s^n$.

Consequently take any $b=rs$ and put $r^n=a+1$ and $s^n=a-1$. What do you can to deduce?

Answer 2

There is a relatively simple elementary proof of this due to E.Z. Chein in the Proceeding of the AMS (from 1976) :

https://www.ams.org/journals/proc/1976-056-01/S0002-9939-1976-0404133-1/S0002-9939-1976-0404133-1.pdf

There are somewhat easier versions of this proof in the literature, if memory serves.

Answer 3

Given $\qquad b^n+1=a^2\implies a^2-b^n-1=0\qquad$ there are at least six solutions for $(a\ne\pm1)$ and an infinite number for $(a=\pm1)$. Here are the indicated solutions given as $(a,b,n)$.

$$(\pm3,2,3),(\pm3,8,1),(\pm2,3,1)\quad \land \quad (\pm1,0,\{1,2,3,...\})$$