## 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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