Find all positive integer values $(x, y, n)$ such that $x^n+1=y^{n+1}$ and $gcd(x, n+1)=1$

Question

My approach to this problem is as follows: First, I attempt to prove that $x^n+1=y^{n+1}$ has solutions on the integers only for $n=1$. Since $y>x$, it follows that $y>1$. Thus If $n\ge2$, we have $$y^{n+1}-x^n>y^n-x^n=(y-x)(y^{n-1}+y^{n-2}x+...+x^{n-1})> 1$$ My reasoning is based on the fact that the factor $(y^{n-1}+y^{n-2}x+...+x^{n-1})$ is at least the sum of $n-1$ powers of $y\geq2$. Therefore, there are no integer solutions for the equation $x^n+1=y^{n+1}$. Now, since $gcd(x,n+1)=1$, it can be concluded that $x$ has to be an odd number, and therefore, $y$ is an even number.

Let $y=2k$ for some integer $k$, then $x=(2k)^2-1=4k^2-1$. That is, the answer to the problem is the triplet $(4k^2-1, 2k, 1)$.

EDIT: I just realized the sentence

Since $y>x$, it follows that $y>1$

Is not correct, since it is possible for $y$ to be lesser than $x$. Therefore, the whole argument is incorrect. However, I'd like to find a way to prove $y^{n+1}-x^n>1$.

Thanks in advance

Answer 1

One guy trying to prove Fermat’s Last Theorem send a telegram: “We should move $y^{n}$ to the right-hand side. The details will be sent a letter”. So let us follow this advice.

We have $x^n=y^{n+1}-1=(y-1)z$, where $z=y^n+y^{n-1}+\dots+1$. We have $\gcd(y-1,z)=(y-1,n+1)$. Since $(y-1)z=x^n$ is coprime with $n+1$, we have that $\gcd(y-1,z)=1$. It follows $y-1=u^n$ and $z=v^n$ for some positive integers $u$ and $v$. If $n>1$ then $y^n<z<(y+1)^n$, a contradiction. So $n=1$, and this case you already solved.

Answer 2

Here's one problematic case for your argument that $n$ must be $1$: what if $y-x$ is negative? If $x > y$, then $x^n$ and $y^{n+1}$ can still plausibly be close to each other.

In fact, Mihăilescu's theorem is one way to show that there is no solution here, though it's a bit of an overpowered tool. Maybe there's a better way using the fact that the powers are off by $1$ from each other, but I'm not seeing anything immediately.

As for what happens once you have $n=1$ and the equation $x+1 = y^2$ where $\gcd(x,2)=1$; we can easily check that any triple $(x,y,n)$ of the form $(4k^2-1, 2k, 1)$ is a solution, and there is nothing left to do (and no assumptions we are making).