EXPLORATION : CAN WE USE POLYNOMIALS TO LEARN ABOUT NUMBERS?

Use a $1 \leftarrow x$ machine to compute each of the following

a. $\dfrac{x^{2}-1}{x-1}$
b. $\dfrac{x^{3}-1}{x-1}$
c. $\dfrac{x^{6}-1}{x-1}$
d. $\dfrac{x^{10}-1}{x-1}$

Can you now see that $\dfrac{x^{\text{number}}-1}{x-1}$ will always have a nice answer without a remainder?

Another way of saying this is that $x^{\text{number}}-1=\left(x-1\right) \times \left( \text{something}\right)$.

For example, you might have seen from part c) that $x^{6}-1=\left(x-1\right) \times \left( x{5}+x{4}+x{3}+x{2}+x+1\right)$.

This means we can say, for example, that $17^{6}-1$ is sure to be a multiple of $16$!
How? Just choose $x=17$ in this formula to get $17^{6}-1=\left(17-1\right) \times \left( \text{something}\right)=\left(16\right) \times \left( \text{something}\right)$.

e. Explain why $999^{100}-1$ must be a multiple of $998$.
f. Can you explain why $2^{100}-1$ must be a multiple of $3$, and a multiple of $15$, and a multiple of $31$ and a multiple of $1023$? (Hint: $2^{100}= \left( 2^2 \right){50}=4{50}$, and so on.)
g. Is $x^{\text{number}}-1$ always a multiple of $x+1$? Sometimes, at least?
h. The number $2^{100}+1$ is not prime. It is a multiple of $17$? Can you see how to prove this?

EXPLORATION: AN INFINITE ANSWER?

Here is a picture of the very simple polynomial $1$ and the polynomial $1-x$.

Can you compute $\dfrac {1}{1-x}$? Can you interpret the answer?

(We’ll explore this example, and more like it, in the island of infinite sums!)