rational solutions of $r^n-s^n=1$?

Question

How can we treat with the Diophantine equation: For what integer values of $a,b,x,y$ and integer $n>1$ $$ \left(\frac ab\right)^n-\left(\frac xy\right)^n=1? $$ Thanks.

Answer 1

As Wojowu said, when $y,b\ne0$

$\left(\frac{a}{b}\right)^{n}-\left(\frac{x}{y}\right)^{n}=1 \iff a^ny^n=x^nb^n+b^ny^n \iff a^ny^n=b^n(x^n+y^n)$

I'm going to search only solutions where $\operatorname{gcd}(a,b)=\operatorname{gcd}(x,y)=1$.

For $n>2$ according to Fermat's last theorem, it must be $a^ny^n = 0$ or $b^ny^n =0$ or $x^nb^n=0$, but $y,b$ must be $\ne0$, so the only solutions are trivial

• if $n>2$ is odd $S=\{a=b\ne0,x=0,y\ne0\}\cup\{a=0,b\ne0,x=-y\ne0\}$

• if $n>2$ is even $S=\{a=\pm b\ne0,x=0,y\ne0\}$

With $n=2$ there are also many non-trivial solutions using pythagorean triples:

$\begin{cases}a=r^2+s^2\\b=2rs\\x=r^2-s^2\\y=b\end{cases}$

where $r,s \in \mathbb{Z}-\{0\}$