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 size relationship
(A) $a>b>1$
(B) $b>a>1$
(C) $a>1.01$
5. (A).

First, $a>1, b>1$. Otherwise, if $0< a \leq 1$, then $a^{2n} \leq 1 < 3a$, which implies $a>1$, a contradiction; if $0< b \leq 1$, then $b^{2n} \leq 1 < 3a$, which implies $b>1$, a contradiction.
Therefore, both $a$ and $b$ are greater than 1.
On one hand,
$$
a^{2n} - b^{2n} = (a+1)^2 - (b+3a) = a^2 - a - b + 1,
$$

On the other hand,
$$
a^{2n} - b^{2n} = (a-b)\left(a^{2n-1} + a^{2n-2}b + \cdots + b^{2n-1}\right),
$$

Thus,
$$
a^2 - a - b + 1 = (a-b)\left(a^{2n-1} + a^{2n-2}b + \cdots + b^{2n-1}\right),
$$

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

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