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) $a>1,01$

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5. (A).

First, $a>1, b>1$. Otherwise,
if $0<1$, then $a>1$, leading to a contradiction;

if $0<3a>3$, then $b>1$, leading to 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 $\frac{a^{2}-a-b+1}{a-b}$
$=a^{2n-1}+a^{2n-2}b+\cdots+b^{2n-1}>1$.
$\therefore \frac{a^{2}-2a+1}{a-b}>0$, i.e., $\frac{(a-1)^{2}}{a-b}>0$. Hence, $a>b$.
In summary, we have $a>b>1$.