Mathematics Asked by highgardener on November 30, 2020

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?**

Given $qquad b^n+1=a^2implies a^2-b^n-1=0qquad$ there are at least six solutions for $(anepm1)$ and an infinite number for $(a=pm1)$. Here are the indicated solutions given as $(a,b,n)$.

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

Answered by poetasis on November 30, 2020

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

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

Answered by Mike Bennett on November 30, 2020

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?

Answered by Piquito on November 30, 2020

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