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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?

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    $\begingroup$ This is a special case of the (proved) Catalan conjecture: en.wikipedia.org/wiki/Catalan%27s_conjecture $\endgroup$
    – Hw Chu
    Sep 27 '19 at 12:00
  • $\begingroup$ Ah of course! Thanks! $\endgroup$
    – highgardener
    Sep 27 '19 at 12:03
  • $\begingroup$ I was wondering @HwChu if there could be a simple proof for this special case? $\endgroup$
    – highgardener
    Sep 27 '19 at 12:30
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    $\begingroup$ Yes, there is the elemenatry proof by Victor Lebesgue from $1850$ for the case that one exponent is $2$. "‘Sur l’impossibilité en nombres entiers de l’equation $x^m = y^2 + 1$’, Nouv. Ann. Math. 9 (1850). $\endgroup$
    – Dietrich Burde
    Sep 27 '19 at 12:43
  • $\begingroup$ The Lebesgue proof is for the "easier" case and does not extend to the proposed equation $a^2-1=b^n$. $\endgroup$
    – Mike Bennett
    Sep 28 '19 at 22:22
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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.

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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?

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  • $\begingroup$ does not seem right. What if $b^{n-1}=a+1,b=a-1?$ e.g. $a=3,b=2,n=3$ then $a+1=2^2=b^2,$ and $a-1=2=b.$ $\endgroup$
    – 111
    Sep 27 '19 at 12:34
  • $\begingroup$ If $\gcd(a+1,a-1) = 1$ then you can claim $a+1 = r^n$ and $a-1 = s^n$. The only chance $\gcd(a+1,a-1) \neq 1$ is when $a+1$ is even. Maybe you need to take extra care about the divisor $2$, but this should be manageable. $\endgroup$
    – Hw Chu
    Sep 27 '19 at 12:36
  • $\begingroup$ Why not to consider the case $\gcd(a+1,a-2)=2?$ $\endgroup$
    – 111
    Sep 27 '19 at 12:41
  • $\begingroup$ Yes, you need to consider that case. In that case it will be $a+1 = \gamma r^n$, $a-1 = \gamma s^n$, where $\gamma = 2$ or $1/2$. $\endgroup$
    – Hw Chu
    Sep 27 '19 at 12:42
  • $\begingroup$ Sorry can't follow you, $\gamma$ can not be a non integer. Please, see the example I provided in my first comment. $\endgroup$
    – 111
    Sep 27 '19 at 12:47
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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 $(a,b,n)$.

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

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