How to prove that $x^2+1=5^y$ has no positive integer solutions for $y\geq 2$?

Question

I am sure I saw a similar question like this one before but I can't find it now. I tried using "order" but failed. It is quite obvious when $y$ is an even number. The real problem is when $y$ is odd. Is there any easy way to solve this? Thanks.

Answer 1

Not a full answer, but way too long for a comment and might have some useful ideas:

We see that, in complex numbers, we need

$$(x+i)(x-i)=(2+i)^y(2-i)^y.$$

So, we have

$$x+i=\pm (2+i)^m(2-i)^n, x-i=\pm (2+i)^{y-m}(2-i)^{y-n}.$$

So,

$$\gcd(x+i,x-i)=\gcd(x+i,x-i)|2i,$$

but a divisor of it is also

$$(2+i)^{\min(m,y-m)}(2-i)^{\min(n,y-n)};$$

as each of $2+i$ and $2-i$ are coprime to $2i$, we must thus have that $m,n\in\{0,y\}$. Clearly, this implies $(m,n)=(0,y)$ or $(m,n)=(y,0)$. Either way,

$$\Im((2+i)^y)=\pm 1.$$

Let $2+i=z=re^{i\theta}$. Then we have that

$$z^y\pm z\in\mathbb{R}$$

$$z^y\pm z = \bar{z}^y\pm\bar{z}$$

$$z^y-\bar{z}^y = \pm(z-\bar{z})$$

$$\frac{z^y-\bar{z}^y}{z-\bar{z}}=\pm 1.$$

Define

$$a_n=\frac{z^n-\bar{z}^n}{z-\bar{z}}.$$

We have that, as $a_n=Az^n+B\bar{z}^n$ for some complex numbers $A$ and $B$,

$$a_{n+2}-4a_{n+1}+5a_n=0,$$

and $a_0=0$, $a_1=1$. It suffices to show that no element of this sequence with index $>1$ has magnitude $1$.

Another idea is that

$$\text{Im}\left(z^n\right)=r^n\sin(n\theta),$$

so we need

$$\sin(n\theta)=\pm\frac{1}{r^n},$$

which implies the existence of a positive integer $m$ so that

$$|n\theta-m\pi|\leq \frac{2}{r^n}$$

(as $x/2<\sin(x)$ for all $0<x<\pi/4$, say). This implies that

$$\left|\frac{\theta}{\pi}-\frac{m}{n}\right|\leq \frac{2}{\pi(nr^n)}.$$

In other words, $\theta/\pi$ is very well-approximated by rational numbers.