5 For any integer $n(n \geqslant 2)$, the positive numbers $a$ and $b$ that satisfy $a^{n}=a+1, b^{2 n}=b+3 a$ have the following relationship:
A. $a>b>1$
B. $b>a>1$
C. $a>1,01$
$5 \mathrm{~A}$. First, $a>1, b>1$. Otherwise,
if $0<a \leq 1$, then $a^{2 n} \leq 1 < 3 a$, which implies $a^{2 n} - b^{2 n} < 0$, thus $a^{2 n} < b^{2 n}$, leading to $a < 1$, a contradiction;
if $0<b \leq 1$, then $b^{2 n} \leq 1 < 3 a$, which implies $a^{2 n} - b^{2 n} > 0$, thus $a^{2 n} > b^{2 n}$, leading to $b < 1$, a contradiction.
Therefore, both $a$ and $b$ are greater than 1.
On one hand, $\quad a^{2 n}-b^{2 n}=(a+1)^{2}-(b+3 a)=a^{2}-a-b+1$
On the other hand, $a^{2 n}-b^{2 n}=(a-b)\left(a^{2 n-1}+a^{2 n-2} b+\cdots+b^{2 n-1}\right)$
Thus,
$$
\begin{array}{c}
a^{2}-a-b+1 \\
=(a-b)\left(a^{2 n-1}+a^{2 n-2} b+\cdots+b^{2 n-1}\right),
\end{array}
$$

which means
$$
\frac{a^{2}-a-b+1}{a-b}=a^{2 n-1}+a^{2 n-2} b+\cdots+b^{2 n-1}>1 \text {. }
$$

Therefore, $\frac{a^{2}-2 a+1}{a-b}>0$, which is $\frac{(a-1)^{2}}{a-b}>0$. Hence, $a>b$.
In summary, we have $a>b>1$.