# 1983 AHSME Problems/Problem 18

## Problem

Let $f$ be a polynomial function such that, for all real $x$, $f(x^2 + 1) = x^4 + 5x^2 + 3$. For all real $x, f(x^2-1)$ is $\textbf{(A)}\ x^4+5x^2+1\qquad \textbf{(B)}\ x^4+x^2-3\qquad \textbf{(C)}\ x^4-5x^2+1\qquad \textbf{(D)}\ x^4+x^2+3\qquad \textbf{(E)}\ \text{none of these}$

## Solution

Let $y = x^2 + 1$. Then $x^2 = y - 1$, so we can write the given equation as \begin{align*}f(y) &= x^4 + 5x^2 + 3 \\ &= (x^2)^2 + 5x^2 + 3 \\ &= (y - 1)^2 + 5(y - 1) + 3 \\ &= y^2 - 2y + 1 + 5y - 5 + 3 \\ &= y^2 + 3y - 1.\end{align*} Then substituting $x^2 - 1$ for $y$, we get \begin{align*}f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\ &= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\ &= x^4 + x^2 - 3.\end{align*} The answer is therefore $\boxed{\textbf{(B)}}$.

## Solution 2

Let $y=x^2.$ We have that \begin{align*}f(x^2 + 1) &= x^4 + 5x^2 + 3 \\ &= y^2 + 5y + 3 \\ &= y^2 + 5y + 4 - 1 \\ &= (y+1)(y+4) -1 \\ &= (x^2 + 1)(x^2 + 4) - 1\end{align*}

Thus, we have $f(n) = n(n+3)-1.$

If we plug in $n = x^2-1$ we have \begin{align*}f(x^2 - 1) &= (x^2 -1)(x^2 + 2)-1 \\ &= (x^4 + x^2 - 2) - 1 \\ &= \boxed{x^4 + x^2 -3} \end{align*}

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