1. Activity 2-11: Quadratic Reciprocity
Activity 2-11: Quadratic Reciprocity

2. 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).

3. 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)

4. 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.

5. 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.

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

7. 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.

8. 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)

9. 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).

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

11. 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...

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