Diophantine equation $p^{2n}+2=3^m$

Question

Let $p$ be a prime and let $m$, $n$ be positive integers. Consider the equation

$$p^{2n}+2=3^m$$

It is easy to see that $(p,m,n)=(5,3,1)$ is a solution. Are there any other solutions?

Answer 1

This isn't a full answer, but so far I've been able to show that $m$ has to be odd.

Because if $m$ is even, then $\frac{m}{2} \in \Bbb{N}$, and $$p^{2n}+2=3^m$$ $$3^m-p^{2n} = 2$$ $$(3^\frac{m}{2}-p^n)(3^\frac{m}{2}+p^n) = 2$$ Since $\frac{m}{2}$ is a positive whole number, and $3^\frac{m}{2}+p^n > 3^\frac{m}{2}-p^n$, this can only happen when $$\text{I.}\ 3^\frac{m}{2}-p^n = 1 \ \ \text{ and } \\ \text{II.}\ 3^\frac{m}{2}+p^n = 2.$$ From the second equation: $$p^n = 2 - 3^\frac{m}{2}$$ Putting this into the first equation: $$3^{\frac{m}{2}}-2+3^{\frac{m}{2}}=1$$ $$2 \times 3^{\frac{m}{2}} = 3$$ Since the left side is even, and the right side is odd, this equation has no solution, so $m$ has to be odd.

Answer 2

That the more general equation $x^2+2=y^n$ has only the solution corresponding to $x=5$ was proven $75$ years ago by Ljunggren. Such a result follows from applying the primitive divisor theorem of Bilu, Hanrot and Voutier.