Activity 2-11: Quadratic Reciprocity

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 we divide by c. So 122 ≡ 45 (mod 11).

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)  a r ≡ b r (mod c)

If a = b + cd, p = q + ce, then a + p = b + q + (d + e)c. If a = b + cd, p = q + ce, then ap = bq + (qd + be +dec)c. If a = b + cd, then a r = b r + f(b, c, d)c.

One equation that has fascinated mathematicians is x 2 ≡ 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 x 2 ≡ 1 (mod 7) have a solution? x = 1 will do. Does x 2 ≡ 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 x 2 ≡ a (mod 7) is soluble only for a = 1, 2 and 4.

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 mathematician Adrien-Marie Legendre, French, (1752 –1833)

What are the properties of the Legendre Symbol? 1. a ≡ b (mod p)  2. 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 3.

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

if p ≡ 1 (mod 4) or if q ≡ (mod 4) if p ≡ q ≡ 3 (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 19 th century. If p and q are distinct odd primes, then

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

Or to do this a different way - Check: 7 2 = 49 = 18 (mod 31).

Task: find. (397 and 757 are both prime.)

297 2 = 397 (mod 757), so the answer IS 1...

With thanks to: The Open University. Carom is written by Jonny Griffiths,