Local olympiad

A little algebra shows that the given polynomial can be written as $(a^2-2a+9)^2-63$ . Letting $n=a^2-2a+9$ , we then need to find integral solutions to the equation $n^2-63=k^2$ ; and while we eventually will need to consider negative solutions for n, those solutions will be just the opposites of any positive solutions we find, so for the moment we restrict ourselves to $n,k \in \mathbb{N}$ .

$\begin{aligned} n^2-63&=k^2 \\ n^2-k^2&=63 \\ (n+k)(n-k)&=63 \end{aligned}$

Since both variables are positive integers, $n+k > n-k$ , so the only possible solutions for $(n+k,n-k)$ are $(63,1), (21,3)$ and $(9,7)$ . Solving these systems for $n$ (and $k$ ), we get the solutions $n=32, 12, 8$ (and $k=31, 9, 1$ respectively.)

Now including the possible negative values for $n$ , we have to solve the quadratic $a^2-2a+9=n$ for $n=\pm8, \pm12, \pm32$ . A little more algebra shows that the only equations that yield integer solutions for $a$ are $n=8, 12$ .

$\begin{aligned} \bullet \ n&=8: &&a^2-2a+9=8 &\Rightarrow \qquad a^2-2a+1=0 &&\Rightarrow \qquad &a=1 \\ \bullet \ n&=12: &&a^2-2a+9=12 &\Rightarrow \qquad a^2-2a-3=0 &&\Rightarrow \qquad &a=-1, 3 \end{aligned}$

Verifying these values, we have

$\begin{aligned} \bullet \ a&=-1: &&a^4-4a^3+22a^2-36a+18 = 81 \\ \bullet \ a&=1: &&a^4-4a^3+22a^2-36a+18 = 1 \\ \bullet \ a&=3: &&a^4-4a^3+22a^2-36a+18 = 81 \end{aligned}$

Thus the only integer values of a for which the given polynomial is a square are $-1, 1, 3$ and the required product is $-1\times 1\times 3 = \boxed{-3}$