Download Prime Numbers - Math Talent, Math Talent Quest Home

Math Talent Quest
Prime Numbers
Number Theory-deals with using the traits of numbers to solve problems. Common topics
related to in number theory include primes, divisibility, mods, and powers.
Primes-Numbers with only divisors of 1 and the number itself.
All numbers can be expressed as a number of primes multiplied together. Being able to analyze
the components of a number can be important in number theory.
Prove: There are an infinite number of primes.
Solution: First, let us assume that there are a finite number of primes. Let these primes be
𝑝1 , 𝑝2 , 𝑝3…𝑝𝑛 . Now, let 𝐴 = 𝑝1 𝑝2 𝑝3..𝑝𝑛 . Then, because all primes are greater than 1, we have
that 𝐴 + ! must be prime, as it is not divisible by any of the primes. Thus, we have a
contradiction, and there must be an infinite number of primes.
Composition-By being able to find a way to factor something, we are able to prove something is
composite. It can also allow simplification of a problem, and can easily lead to a solution. For
example while manually solving 712 − 292 would be very annoying, we can easily factor it into
(71 + 29)(71 − 29) = (100)(42) = 4200
Simple Factorings: 𝑥 2 + 2𝑥𝑦 + 𝑦 2 = (𝑥 + 𝑦)2, 𝑥 2 − 2𝑥𝑦 + 𝑦 2 = (𝑥 − 𝑦)2, 𝑥 2 − 𝑦 2 =
(𝑥 + 𝑦)(𝑥 − 𝑦)
SFFT, or Simon’s Favorite Factoring Trick has 𝑥𝑦 + 𝑥 + 𝑦 + 1 = (𝑥 + 1)(𝑦 + 1)
Factoring cubes: 𝑥 3 + 𝑦 3 = (𝑥 + 𝑦)(𝑥 2 − 𝑥𝑦 + 𝑦 2 ), 𝑥 3 − 𝑦 3 = (𝑥 − 𝑦)(𝑥 2 + 𝑥𝑦 + 𝑦 2 )
Example 1: Prove that 2𝑎2 𝑏 − 𝑎2 − 2𝑏 + 1 is composite for all 𝑎 and 𝑏 ≥ 2
Solution: Let’s use SFFT. 𝑥𝑦 + 𝑥 + 𝑦 + 1 = (𝑥 + 1)(𝑦 + 1), and let us have 𝑥 = −𝑎2 and
have that 𝑦 = −2𝑏. Then we have 2𝑎2 𝑏 − 𝑎2 − 2𝑏 + 1 = (−𝑎2 + 1)(−2𝑏 + 1) =
(𝑎2 − 1)(2𝑏 − 1). Because we have that 𝑎 and 𝑏 are greater than or equal to 2, each of 𝑎2 − 1
and 2𝑏 − 1 cannot be less than 3, so together they must make a composite number.
Example 2: Find the value of the largest prime factor of 216 − 29 + 1
Solution: We could calculate this out, but it would not be pretty. Instead, let’s factor! We can
group this as (28 )2 − 2(28 ) + 1 = (28 − 1)2 = [(24 + 1)(24 − 1)]2 From here, it is clear the
largest prime factor is 24 + 1 = 17
Fermat’s Little Theorem- states that if 𝑎 is an integer, 𝑝 is a prime number, and 𝑎 is not
divisible by 𝑝, then 𝑎𝑝−1 ≡ 1 (mod 𝑝).
Proof of Fermat’s Little Theorem: Let us have a necklace with 𝑝 beads, and each bead can be
one of 𝑎 colors. Thus, there are 𝑎𝑝 ways to color the necklace, with 𝑎 necklaces of the same
color. For the other necklaces, there are 𝑝 ways to rotate the necklace and get the same coloring.
Thus, we have that 𝑎𝑝 − 𝑎 must be divisible by 𝑝, and 𝑎𝑝 ≡ 𝑎(mod)𝑝, or that 𝑎𝑝−1 ≡ 1 (mod 𝑝).
Example 3: Find 440 + 550 + 660 mod 7
Solution:
Because 7 is prime, and 4,5,6 not divisible by 7, we can use Fermat’s Little Theorem. From
there we get 46 ≡ 56 ≡ 66 ≡ 1 mod 7. Thus, we get that 44 + 52 + 1 = 440 + 550 + 660 mod 7,
and we easily find 44 + 52 + 1 = 256 + 25 + 1 = 282 ≡ 2 mod 7.
Example 4:
Finally, let us prove Wilson’s Theorem.
Prove: If 𝑝 is an integer, (𝑝 − 1)! + 1 is a multiple of 𝑝 if and only if 𝑝 is prime.
Solution:
First, assume 𝑝 is composite. Then 𝑝 has a factor 𝑑 > 1 that is less than 𝑝 − 1. Then 𝑑 divides
(𝑝 − 1)!, so 𝑑 does not divide (𝑝 − 1)! + 1.
Suppose 𝑝 is prime. For every integer 1,2…𝑝 − 1 there is an inverse mod 𝑝, such that 𝑎𝑏 ≡ 1
mod 𝑝, and with one number such that it is its own inverse, and must be −1 mod 𝑝. Thus, by
multiplying all of these pairs together we get (𝑝 − 1)! ≡ −1 mod 𝑝.