1
$\begingroup$

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?

$\endgroup$
5
  • 4
    $\begingroup$ This is a special case of the (proved) Catalan conjecture: en.wikipedia.org/wiki/Catalan%27s_conjecture $\endgroup$
    – Hw Chu
    Sep 27, 2019 at 12:00
  • $\begingroup$ Ah of course! Thanks! $\endgroup$
    – highgardener
    Sep 27, 2019 at 12:03
  • $\begingroup$ I was wondering @HwChu if there could be a simple proof for this special case? $\endgroup$
    – highgardener
    Sep 27, 2019 at 12:30
  • 2
    $\begingroup$ 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, 2019 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, 2019 at 22:22

3 Answers 3

Reset to default
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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?

$\endgroup$
7
  • $\begingroup$ 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, 2019 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, 2019 at 12:36
  • $\begingroup$ Why not to consider the case $\gcd(a+1,a-2)=2?$ $\endgroup$
    – 111
    Sep 27, 2019 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, 2019 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, 2019 at 12:47
0
$\begingroup$

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 $(a,b,n)$.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.