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

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## Contents |

## Project -- The Square Root of -1[edit]

**Notation:** In modular arithmetic, if

for some *m*, then we can write

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

Note that if *x* satisfies *x*^{2} ≡ *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

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

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

have a solution? Why?

5. If *p* ≡ 1 (mod 4) prime and *q* ≡ 1 (mod 4) prime and p ≠ q. Show that

has 4 solutions.

6. Find the 4 solutions to

note that 493 = 17 × 29.

7. Take an integer *n* with more than 2 prime factors. Consider:

Under what condition is there a solution? Explain thoroughly.