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

High School Mathematics Extensions

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Supplementary Chapters — Primes and Modular Arithmetic — Logic

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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 ≡ y, then m - x ≡ -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 ≡ 1 (mod 4) prime. Explain why no square root of -1 exist if p ≡ 3 (mod 4) prime.

2. Show that for p ≡ 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 ≡ a (mod m)

x ≡ b (mod n)

is that number. E.g. Suppose m = 2, n = 3, then 4 is uniquely represented by

x ≡ 0 (mod 2)

x ≡ 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 ≡ 1 (mod 4) prime and q ≡ 3 (mod 4) prime. Does

$x\equiv {\sqrt {-1}}{\pmod {pq}}$

have a solution? Why?

5. If p ≡ 1 (mod 4) prime and q ≡ 1 (mod 4) prime and p ≠ 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 × 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.