Let $A=(3+\sqrt{7})^n$ and $B=(3-\sqrt{7})^n$. Then by binomial expansion, we know the value of $(A+B)$ must be an even integer because all the odd power of $\sqrt{7}$ are canceled and the rest terms are all integers and doubled. $$\begin{array}{rl} A + B &= (3+\sqrt{7})^n + (3-\sqrt{7})^n\\ &= (3^n + \binom{n}{1}\cdot 3^{n-1}\cdot\sqrt{7} + \binom{n}{2}\cdot 3^{n-2}\cdot(\sqrt{7})^2+\cdots)\\ &+ (3^n - \binom{n}{1}\cdot 3^{n-1}\cdot\sqrt{7} + \binom{n}{2}\cdot 3^{n-2}\cdot(\sqrt{7})^2+\cdots) \\ &= 2\times\left(3^n + \binom{n}{2}\cdot 3^{n-2}\cdot 7 + \cdots\right) \end{array}$$

Meanwhile, because $0 < 3-\sqrt{7} < 1$, we find $0 < (3-\sqrt{7})^n < 1$. Therefore, $X = (A+B)-1$ which means $X$ is odd.