Given $x_m=4x_{m-1}-x_{m-2},\ x_1=1,\ x_2=3$

Question

I friend told me that apart from trivial ones, the elements in this sequence never equal powers of 3: $$x_m=4x_{m-1}-x_{m-2},\ x_1=1,\ x_2=3.$$

Could you please help me to prove this?

Answer 1

Hint: We have: $$ x_m = \frac{1}{6}\left[(3-\sqrt{3})(2+\sqrt{3})^n+(3+\sqrt{3})(2-\sqrt{3})^n\right]$$ and we may notice that $3x_m^2-2$ is always a square, so it is enough to prove that $$ 3^{2k+1}-2 = a^2 \tag{1}$$ has the only solutions $(k,a)\in\{(0,1),(1,5)\}$. $(1)$ is equivalent to: $$ 3 (3^k-1)(3^k+1) = (a-1)(a+1) \tag{2} $$ that is not difficult to study.