## The Astounding Power of Area

### 2.4 Don’t Forget: \(x\) can actually be a number!

In the previous lesson we saw how to divide polynomials – at least ones that divide nicely. (We’ll deal with remainders in the next lesson.)

This side lesson gives a little taste of the power of polynomial work in Number Theory.

**EXERCISE 1:**

*a) Use the reverse area method to evaluate \(\dfrac{x^{3}+8}{x+2}\).*

*b) Making this a little more abstract, evaluate \(\dfrac{x^{3}+a^{3}}{x+a}\).*

*c) Use the reverse area method to show that \(x^{5}+a^{5}\) is evenly divisible by \(x+a\).*

*d) Explain why, for \(n\) odd, \(x^{n}+a^{n}\) is a multiple of \(x+a\).*

*e) Show that \(7^{50}+22^{25}\) is divisible by \(71\).*

**EXERCISE 2: **

*a) Show that \(x^{n}-1=\left(x-1\right)\times\left( some \; polynomial\right)\) for \(n\) a positive integer.*

This shows, for example, that the number \(20^{463}-1\) is divisible by \(19\) and is not prime.

*b) List three factors of \(4^{20}-1\) different from \(1\) and the number itself.*

*c) Show that \(2^{n}-1\) cannot be prime if \(n\) is a composite number.*

Part c) shows that \(2^{composite}-1=composite\). We wonder now if \(2^{prime}-1\) is prime? It is certainly the case for the first few cases:

\(2^{2}-1=3\), prime.

\(2^{3}-1=7\), prime.

\(2^{5}-1=31\), prime.

\(2^{7}-1=127\), prime.

\(2^{11}-1=2047\), prime.

Prime numbers of this form are today called *Mersenne primes*, in honor of French Monk Marin Mersenne (1588-1648). He was interested in these prime for his work on perfect numbers. (See the video on perfect numbers.)

*d) (INTERNET RESEARCH): Is \(2^{prime}-1\) always a prime number?*

*e) (INTERNET RESEARCH): What is the largest Mersenne prime known today?*

“Amateur” mathematician Pierre de Fermat (1601-1655) examined numbers of the form \(2^{n}+1\).

*f) Show that if \(n\) has an odd factor greater than one, then \(2^{n}+1\) cannot be prime. (Find a general formula for \(\dfrac{x^{n}+1}{x+1}\) for \(n\) odd.)*

The only numbers possessing no odd factors greater than one are the powers of two: \(1, 2, 4, 8, 16, 32, \ldots\). The question now is: *Are numbers of the form \(2^{2^{k}}+1\) sure to be prime?* The first few certainly are.

*\(2^{1}+1=3\), prime.*

*\(2^{2}+1=5\), prime.*

*\(2^{4}+1=17\), prime.*

*\(2^{8}+1=257\), prime.*

*g) (INTERNET RESEARCH): What did Leonhard Euler (1707-1783) discover about Fermat primes?*

*h) (INTERNET RESEARCH): What is the largest Fermat prime known today?*

*i) (INTERNET RESEARCH:) What is the largest pair of “twin primes” known today?*

**EXERCISE 3: ***Suppose \(P\) is a polynomial with integer coefficients. *

*a) Explain why \(P\left(x\right)-P\left(a\right)\) is sure to be a multiple of \(x-a\).*

*b) Suppose \(P\left(a\right) = 19\), \(P\left(b\right) = 20\), and \(P\left(c\right) = 21\) for three distinct integers \(a\), \(b\), and \(c\). Explain why \(a\), \(b\), and \(c\) must be consecutive integers. *

**SOME SOLUTIONS**

1. a) *\(\dfrac{x^{3}+8}{x+2}=x^{2}-2x+4\)*

d) \(\dfrac{x^{n}+a^{n}}{x+a}=x^{n-1}-ax^{n-2}+a^{2}x^{n-3}-a^{3}x^{n-4}+\ldots-a^{n-2}x+a^{n-1}\).

e) \(49^{25}+22^{25}\) is divisible by \(49+22=71\).

**2.** a) \(\dfrac{x^{n}-1}{x-1}=x^{n-1}+x^{n-2}+\ldots +x+1\).

b) \(4^{20}-1=3\times something\); \(16^{10}-1=15 \times somthing\) (and so \(5\) is yet another factor.); \(256^{5}-1=255 \times something\)

c) If \(n=ab\), then \(2^{n}-1=\left(2^{b}\right)^{a}-1=\left(2^{b}-1\right) \times something\).

If \(b>1\), then \(2^{b}-1\) is a factor different from \(1\).

f) \(\dfrac{x^{n}+1}{x+1}=x^{n-1}-x^{n-2}+x^{n-3}-x^{n-4}+\ldots-x+1\) for \(n\) odd. So \(x^{n}+1\) is a multiple of \(x+1\) if \(n\) is odd.

If \(n=ab\) with \(b\) an odd factor, then \(2^{n}+1=\left(2^{a}\right)^{b}+1\) is a multiple of \(2^{a}+1\) and so is not prime.

**3.** a) \(\dfrac{x^{n}-a^{n}}{x-a}=x^{n-1}+ax^{n-2}+a^{2}x^{n-3}+\ldots+a^{n-2}x+a^{n-1}\). So \(x^{n}-a^{n}\) is always a multiple of \(x-a\). If \(P\left(x\right) = b_{n}x^{n}+b_{n-1}x^{n-1}+\ldots b_{1}x+b_{0}\), then \(P\left(x\right) – P\left(a\right) = b_{n}\left(x^{n}-a^{n}\right)+b_{n-1}\left(x^{n-1}-a^{n-1}\right) + \ldots +b_{1}\left(x-a\right)\). This is a multiple of \(x-a\).

b) \(P\left(b\right)-P\left(a\right)\) is a multiple of \(b-a\). But \(P\left(b\right)-P\left(a\right)=20-19=1\), so \(a\) and \(b\) differ by \(1\). Similarly, \(b\) and \(c\) differ by \(1\). So \(a\), \(b\), and \(c\) are consecutive.

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