[Algebra]
Problem: Frank Narf accidentally read a degree $n$ polynomial with integer coefficients backwards. That is, he read $a_{n} x^{n}+\ldots+a_{1} x+a_{0}$ as $a_{0} x^{n}+\ldots+a_{n-1} x+a_{n}$. Luckily, the reversed polynomial had the same zeros as the original polynomial. All the reversed polynomial's zeros were real, and also integers. If $1 \leq n \leq 7$, compute the number of such polynomials such that $\operatorname{GCD}\left(a_{0}, a_{1}, \ldots, a_{n}\right)=1$.

Solution: When the coefficients of a polynomial $f$ are reversed to form a new polynomial $g$, the zeros of $g$ are the reciprocals of the zeros of $f: r$ is a zero of $f$ if and only if $r^{-1}$ is a zero of $g$. In this case, the two polynomials have the same zeros; that is, whenever $r$ is a zero of either, so must be $r^{-1}$. Furthermore, both $r$ and $r^{-1}$ must be real as well as integers, so $r= \pm 1$. As the only zeros are \pm 1 , and the greatest common divisor of all the coefficients is 1 , the polynomial must have leading coefficient 1 or -1 . Thus

$$
\begin{aligned}
f(x) & = \pm(x \pm 1)(x \pm 1) \cdots(x \pm 1) \\
& = \pm(x+1)^{k}(x-1)^{n-k}
\end{aligned}
$$

If $A_{n}$ is the number of such degree $n$ polynomials, then there are $n+1$ choices for $k, 0 \leq k \leq n$. Thus $A_{n}=2(n+1)$. The number of such degree $n$ polynomials for $1 \leq n \leq 7$ is the sum:

$$
A_{1}+A_{2}+\ldots+A_{7}=2(2+3+\ldots+8)=2 \cdot 35=\mathbf{7 0}
$$

Final answer: 70