# Solutions to the Diophantine equation

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) :

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.

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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