7.114 Let $f$ be a polynomial function such that for all real values of $x$, $f\left(x^{2}+\right.$ 1) $=x^{4}+5 x^{2}+3$. Then for all real values of $x, f\left(x^{2}-1\right)$ is
(A) $x^{4}+5 x^{2}+1$.
(B) $x^{4}+x^{2}-3$.
(C) $x^{4}-5 x^{2}+1$.
(D) $x^{4}+x^{2}+3$.
(E) None of the above.
(34th American High School Mathematics Examination, 1983)
[Solution 1] Since $f(x)$ is a polynomial function, and the degree of $f\left(x^{2}+1\right)$ is 4, the degree of $f(x)$ is 2.
Let $f(x)=a x^{2}+b x+c$, where $a, b, c$ are constants, then we have
$$
\begin{aligned}
f\left(x^{2}+1\right) & =x^{4}+5 x^{2}+3 \\
& =a\left(x^{2}+1\right)^{2}+b\left(x^{2}+1\right)+c \\
= & a x^{4}+(2 a+b) x^{2}+(a+b+c),
\end{aligned}
$$

We have $a=1, 2a+b=5$ and $a+b+c=3$.
Solving these, we get $a=1, b=3, c=-1$,
which gives $f(x)=x^{2}+3 x-1$.

Thus,
$$
\begin{aligned}
f\left(x^{2}-1\right) & =\left(x^{2}-1\right)^{2}+3\left(x^{2}-1\right)-1 \\
& =x^{4}-2 x^{2}+1+3 x^{2}-3-1 \\
& =x^{4}+x^{2}-3 .
\end{aligned}
$$

Therefore, the answer is $(B)$.
[Solution 2] Write $x^{4}+5 x^{2}+3$ in the form of powers of $x^{2}+1$:
$$
\begin{aligned}
x^{4}+5 x^{2}+3 & =\left(x^{4}+2 x^{2}+1\right)+\left(3 x^{2}+3\right)-1 \\
& =\left(x^{2}+1\right)^{2}+3\left(x^{2}+1\right)-1 .
\end{aligned}
$$

Let $w=x^{2}+1$, then $f(w)=w^{2}+3 w-1$. Therefore, for all $x$, we have
$$
\begin{aligned}
f\left(x^{2}-1\right) & =\left(x^{2}-1\right)^{2}+3\left(x^{2}-1\right)-1 \\
& =x^{4}-2 x^{2}+1+3 x^{2}-3-1 \\
& =x^{4}+x^{2}-3 .
\end{aligned}
$$

Therefore, the answer is $(B)$.