Solving for Primes and Evens: $x^n+x^{n-1}+\cdots+x+1=y^2+y+1$

[anemone]

Find all primes $x$ and $y$ and even numbers $n>2$ satisfying the equation $x^n+x^{n-1}+\cdots+x+1=y^2+y+1$.

 

[Nono713]

Spoiler

First of all, note that since $n > 2$ we must have $x \ne y$ since equality cannot be achieved if $x = y$. Now let $x, y$ be distinct primes and $n > 2$ be an even integer such that the equation holds.

<insert proof that $x = 2$ here; my previous proof was flawed>

... therefore $x = 2$. Now observe that:
$$x^n + x^{n - 1} + \cdots + 1 = \frac{x^{n + 1} - 1}{x - 1}$$
Therefore:
$$2^n + 2^{n - 1} + \cdots + 1 = 2^{n + 1} - 1$$
So that we are now solving the following equation for $y$ an odd prime (recall $x \ne y$) and $n > 2$ integer:
$$2^{n + 1} - 1 = y^2 + y + 1 ~ ~ ~ \iff ~ ~ ~ y^2 + y + \left ( 2 - 2^{n + 1} \right ) = 0$$
Using the quadratic equation discarding the negative solution we find:
$$y = \frac{-1 + \sqrt{1 - 4 \left ( 2 - 2^{n + 1} \right )}}{2} = \frac{\sqrt{2^{n + 3} - 7} - 1}{2}$$
We are therefore now solving the diophantine equation in $n > 2$ and $z$ given by:
$$2^{n + 3} - 7 = z^2$$
This is a variant of the Ramanujan-Nagell equation given by $2^m - 7 = x^2$, and its solutions are in one to one correspondence with the solutions to our equation, by a direct change of variables $(m, x) \to (n + 3, z)$. It is known that the Ramanujan-Nagell equation has only five solutions in $m$, given by $m = 3, 4, 5, 7, 15$. These translate to $n = 0, 1, 2, 4, 12$, of which only the solutions $n = 4$ and $n = 12$ are admissible. Working backwards, this gives:
$$n = 4 ~ ~ ~ \implies ~ ~ ~ z = 11 ~ ~ ~ \implies ~ ~ ~ y = 5$$
$$n = 12 ~ ~ ~ \implies ~ ~ ~ z = 181 ~ ~ ~ \implies ~ ~ ~ y = 90$$
But $y = 90$ isn't prime, and so does not satisfy the conditions, and we therefore conclude that the only solution $(x, y, n)$ is $(2, 5, 4)$.

---

Okay, so using Ramanujan-Nagell was a bit of a cop-out, I guess. I have a proof of it in front of me and it's not what I would call elementary, and I'd be interested to see solutions that don't need to invoke it, especially since I didn't manage to prove that there are no other solutions with $x \ne 2$; perhaps the restriction to $y$ being a prime is essential to solving the challenge efficiently... I also just noted that it's easy to show that $n$ must divide $y - 1$, which may hint at an alternative proof!

 

[anemone]

Thanks, Thomas for your well written solution!:)

I will show an elementary method (provided by other) to solve for this challenge here:

Spoiler

Obviously $x\ne y$.

Rewrite the given equation as

$x(x^{n-1}+x^{n-2}+\cdots+1)=y(y+1)$

If $y\le x^{\tiny{\dfrac{n}{2}}}-1$, then $y<x^{\tiny{\dfrac{n}{2}}}$ and hence we see that $y^2<x^n$.

By adding both inequalities, we obtain:

$y^2+y<x^n+x^{\tiny{\dfrac{n}{2}}}<x^n+x^{n-1}+\cdots+x$ since $n>2$.

It follows that $y\ge x^{\tiny{\dfrac{n}{2}}}$.

Since $n>2$ and is an even number, $\dfrac{n}{2}$ is a natural number larger than 1. This implies that $y\ne x^{\tiny{\dfrac{n}{2}}}$ by the given condition that $y$ is a prime.

We conclude that $y\ge x^{\tiny{\dfrac{n}{2}}}+1$.

We may also write the given equation as

$x(x^{\tiny{\dfrac{n}{2}}}-1)(x^{\tiny{\dfrac{n}{2}}}+1)=(x-1)y(y+1)$

This shows that $y$ divides $(x^{\tiny{\dfrac{n}{2}}}-1)(x^{\tiny{\dfrac{n}{2}}}+1)$. But $y\ge x^{\tiny{\dfrac{n}{2}}}+1$ and $y$ is a prime. Hence the only possibility is $y=x^{\tiny{\dfrac{n}{2}}}+1$.

This gives

$x(x^{\tiny{\dfrac{n}{2}}}-1)=(x-1)(y+1)=(x-1)(x^{\tiny{\dfrac{n}{2}}}+2)$

Simplification leads to $3x=x^{\tiny{\dfrac{n}{2}}}+2$. This shows that $x$ divides 2. Thus, $x=2$ and hence $y=5$, $n=4$.