(a) Factor $x^{2}-4$ as the product of two linear expressions in $x$.

(b) Determine the integer $k$ for which $98^{2}-4=100 k$.

(c) Determine the positive integer $n$ for which $(20-n)(20+n)=391$.

(d) Prove that 3999991 is not a prime number. (A prime number is a positive integer greater than 1 whose only positive divisors are 1 and itself. For example, 7 is a prime number.)
(a) The expression $x^{2}-4$ is a difference of squares which can be factored as

$$
x^{2}-4=(x+2)(x-2)
$$

(b) Using (a) with $x=98$, we obtain $98^{2}-4=(98+2)(98-2)=100 \cdot 96$ and so $k=96$.

Alternatively, $98^{2}-4=9604-4=9600=100 \cdot 96$ which gives $k=96$.

(c) If $(20-n)(20+n)=391$, then $20^{2}-n^{2}=391$.

This gives $400-n^{2}=391$ from which we obtain $n^{2}=9$ and so $n= \pm 3$.

Since $n$ is positive, then $n=3$.

We can verify that $17 \cdot 23=391$.

(d) We note that $3999991=4000000-9=2000^{2}-3^{2}=(2000+3)(2000-3)=2003 \cdot 1997$. Since we have written 3999991 as the product of two positive integers that are each greater than 1, we can conclude that 3999991 is not a prime number.