Prove that $x^2 + 8 = 3^y$ has only one solution: $x=1, y=2,$ where $x,y\in \mathbb{N}$

Question

So I've been practicing some Diophantine equations, but this is the first one where the power is a variable, I don't even know how to begin.

Prove that $x^2 + 8 = 3^y$ has only one solution: $x=1, \ y=2,$ where $x,y\in \mathbb{N}$

Answer 1

Hint:

As $x$ is a solution, so will be $-x\implies$ WLOG $x\ge0$

Clearly, $x$ must be odd (why?)

$\implies x^2+8\equiv1\pmod8\implies y$ must be even $=2z$(say)

$$8=3^{2z}-x^2=(3^z-x)(3^z+x)$$

$$\iff\dfrac{3^z+x}2\cdot\dfrac{3^z-x}2=2\cdot1=-2\cdot-1$$

Now $\dfrac{3^z+x}2\ge\dfrac{3^z-x}2$ as $x\ge0$

and $\dfrac{3^z+x}2>0$

Can you take it from here?

Answer 2

Welcome to MSE: another idea simpler but longer. $$x^2 + 8 = 3^y\\x^2-1=3^y-9\\(x-1)(x+1)=9(3^{y-2}-1)$$so check all possibilities $$\underbrace{(x-1)}_{0,1,3,9}(x+1)=3.3.\underbrace{(3^{y-2}-1)}_{}\\\to x-1=0\to (3^{y-2}-1)=0 \to y=2\\\to x-1=1\to 3^y=4+8\to y\not \in \mathbb{N}\\\to x-1=3\to 3^y=16+8\to y\not \in \mathbb{N}\\ \to x-1=9\to 3^y=100+8\to y\not \in \mathbb{N}$$