Activity 2-11: Quadratic Reciprocity www.carom-maths.co.uk Activity 2-11: Quadratic Reciprocity

both a and b have the same remainder when divide by c. Modulo arithmetic is a helpful way to look at the properties of numbers. We say a ≡ b (mod c) if c divides a - b. Another way of putting it is that both a and b have the same remainder when divide by c. So 122 ≡ 45 (mod 11).

a ≡ b (mod c)  ar ≡ br (mod c) Task: can you prove the following? a ≡ b (mod c) and p ≡ q (mod c)  a + p ≡ b + q (mod c) a ≡ b (mod c) and p ≡ q (mod c)  ap ≡ bq (mod c) a ≡ b (mod c)  ar ≡ br (mod c)

One equation that has fascinated mathematicians is x2 ≡ a (mod p). where p is prime, and a is not divisible by p. For what values of a does this have solutions? Let’s try p = 7. Does x2 ≡ 1 (mod 7) have a solution? x = 1 will do. Does x2 ≡ a (mod 7) have a solution for other values of a? Trying out x = 2, 3, 4, 5, 6 yields the values 4, 2, 2, 4, 1 (mod 7). So the equation x2 ≡ a (mod 7) is soluble only for a = 1, 2 and 4.

which of the numbers from 1 to p - 1 are which. We call the values 1, 2 and 4 the quadratic residues of 7, while 3, 5 and 6 are the quadratic non-residues of 7. It turns out that for any odd prime p, there are always quadratic residues, and non-residues. It would be wonderful if we could find a way of discerning which of the numbers from 1 to p - 1 are which.

Introducing... the Legendre Symbol. The only surviving portrait(!) of the French mathematician Adrien-Marie Legendre (1752 –1833)

If p is an odd prime and a is not divisible by p, then   1 if a is a quadratic residue of p -1 if a is a quadratic non-residue of p What are the properties of the Legendre Symbol? 1. a ≡ b (mod p)  2. 3.

We also have that 1 if p ≡ 1 (mod 4) -1 if p ≡ 3 (mod 4) 1 if p ≡ 1 (mod 8) or if p ≡ 7 (mod 8) -1 if p ≡ 3 (mod 8) or if p ≡ 5 (mod 8)

if p ≡ 1 (mod 4) or if q ≡ (mod 4) We can add to these the Law of Quadratic Reciprocity, seen as one of the most beautiful jewels of number theory in the 19th century. If p and q are distinct odd primes, then if p ≡ 1 (mod 4) or if q ≡ (mod 4) if p ≡ q ≡ 3 (mod 4).

Combining these rules together should hopefully give us enough to find out for all odd primes p. Task: find

Or to do this a different way - Check: 72 = 49 = 18 (mod 31). Task: find . (397 and 757 are both prime.) 2972 = 397 (mod 757) , so the answer IS 1...

With thanks to: The Open University. Carom is written by Jonny Griffiths, mail@jonny-griffiths.net