Solutions to the Diophantine equation

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?

• This is a special case of the (proved) Catalan conjecture: en.wikipedia.org/wiki/Catalan%27s_conjecture Sep 27, 2019 at 12:00
• Ah of course! Thanks! Sep 27, 2019 at 12:03
• I was wondering @HwChu if there could be a simple proof for this special case? Sep 27, 2019 at 12:30
• 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). Sep 27, 2019 at 12:43
• The Lebesgue proof is for the "easier" case and does not extend to the proposed equation $a^2-1=b^n$. Sep 28, 2019 at 22:22

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.

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?

• 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 $a-1=2=b.$
– 111
Sep 27, 2019 at 12:34
• 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. Sep 27, 2019 at 12:36
• Why not to consider the case $\gcd(a+1,a-2)=2?$
– 111
Sep 27, 2019 at 12:41
• 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 $1/2$. Sep 27, 2019 at 12:42
• Sorry can't follow you, $\gamma$ can not be a non integer. Please, see the example I provided in my first comment.
– 111
Sep 27, 2019 at 12:47

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 $$(a,b,n)$$.

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