Teacher Lorena taught her students the following notable product: for any real numbers $a$ and $b$,

$$
a^{2}-b^{2}=(a+b)(a-b)
$$

For example, $4^{2}-3^{2}=16-9=7$. On the other hand, $(4+3)(4-3)=7 \times 1=7$. Using this teaching from Teacher Lorena,

a) Calculate

$$
100^{2}-99^{2}+98^{2}-97^{2}+96^{2}-95^{2}+\cdots+2^{2}-1^{2}
$$

b) Find two integers greater than 1 whose product is 999,991.
Solution

a) Using the notable product $a^{2}-b^{2}=(a+b)(a-b)$, we will rewrite the expression

$$
100^{2}-99^{2}+98^{2}-97^{2}+96^{2}-95^{2}+\cdots+2^{2}-1^{2}
$$

as

$$
(100+99)(100-99)+(98+97)(98-97)+(96+95)(96-95)+\ldots+(2+1)(2-1)
$$

Notice that many of the differences above are equal to one! For example, $100-99=1$, $98-97=1$, and so on. Therefore, the expression above is equal to

$$
100+99+98+97+\cdots+3+2+1
$$

To calculate the sum above, we will add it twice! Observe:

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-29.jpg?height=186&width=1067&top_left_y=2286&top_left_x=376)

All the sums are 101. Since there are 100 sums, this gives $100 \times 101=10100$. But we had added what we wanted twice, so we need to divide by two. Thus, the final answer is

$$
\frac{10100}{2}=5050
$$

b) Using the notable product taught by Professor Lorena again, we observe that

$$