High School Mathematics Extensions/Primes/Project/The Square Root of -1

Project -- The Square Root of -1
Notation: In modular arithmetic, if
 * $$x^2 \equiv y \pmod{m} \!$$

for some m, then we can write
 * $$x \equiv \sqrt{y} \pmod{m}$$

we say, x is the square root of y mod m.

Note that if x satisfies x2 &equiv; y, then m - x &equiv; -x when squared is also equivalent to y. We consider both x and -x to be square roots of y.

1. Question 5 of the Problem Set showed that
 * $$x \equiv \sqrt{-1} \equiv \sqrt{p-1} \pmod{p}$$

exists for p &equiv; 1 (mod 4) prime. Explain why no square root of -1 exist if p &equiv; 3 (mod 4) prime.

2. Show that for p &equiv; 1 (mod 4) prime, there are exactly 2 solutions to
 * $$x \equiv \sqrt{-1} \pmod{p}$$

3. Suppose m and n are integers with gcd(n,m) = 1. Show that for each of the numbers 0, 1, 2, 3, .... , nm - 1 there is a unique pair of numbers a and b such that the smallest number x that satisfies:
 * x &equiv; a (mod m)
 * x &equiv; b (mod n)

is that number. E.g. Suppose m = 2, n = 3, then 4 is uniquely represented by
 * x &equiv; 0 (mod 2)
 * x &equiv; 1 (mod 3)

as the smallest x that satisfies the above two congruencies is 4. In this case the unique pair of numbers are 0 and 1.

4. If p &equiv; 1 (mod 4) prime and q &equiv; 3 (mod 4) prime. Does
 * $$x \equiv \sqrt{-1} \pmod{pq}$$

have a solution? Why?

5. If p &equiv; 1 (mod 4) prime and q &equiv; 1 (mod 4) prime and p &ne; q. Show that
 * $$x \equiv \sqrt{-1} \pmod{pq}$$

has 4 solutions.

6. Find the 4 solutions to
 * $$x \equiv \sqrt{-1} \pmod{493} $$

note that 493 = 17 &times; 29.

7. Take an integer n with more than 2 prime factors. Consider:
 * $$x \equiv \sqrt{-1} \pmod{n}$$

Under what condition is there a solution? Explain thoroughly.