Problem (Number Theory):
Compute the least integer value of the function

$$
f(x)=\frac{x^{4}-6 x^{3}+2 x^{2}-6 x+2}{x^{2}+1}
$$

whose domain is the set of all real numbers.

Solution:
$\quad$ Use polynomial long division to rewrite $f(x)$ as

$$
f(x)=x^{2}-6 x+1+\frac{1}{x^{2}+1}
$$

The quadratic function $x^{2}-6 x+1=(x-3)^{2}-8$ has a minimum of -8 , achieved at $x=3$. The "remainder term" $\frac{1}{x^{2}+1}$ is always positive. Thus $f(x)>-8$ for all $x$, so any integer value of $f(x)$ must be at least -7 .

When $x=3$, the remainder term is less than 1 , so $f(3)$ is less than -7 . But $f(4)=-\frac{34}{5}>-7$, so there must be some value of $x$ between 3 and 4 for which $f(x)=-7$, so the least integer value of $f(x)$ is $\mathbf{- 7}$. The reader may note that $f(x)=-7$ when $x \approx 2.097$ and $x \approx 3.970$.

Final answer: -7