SPLASH! 2012 QUADRATIC RECIPROCITY Michael Belland

 Thank you for taking “Is the remainder a square? Elementary Number Theory and Quadratic Reciprocity” at Splash!  This powerpoint is intended to provide a review of what we covered in class. Another powerpoint coming soon will prove the statement of quadratic reciprocity.  If you read this powerpoint, feel free to take breaks once in a while! There’s a lot of content in here. It took me six weeks of dedicated study to really understand all of this. If you have any questions, please me! Introduction

 Modular Arithmetic  Sigma and Pi Notation  Floor function  Why should we care about Quadratic Reciprocity?  Legendre Symbols A Brief Outline

 “Clock Arithmetic”  For example, 6 hours after 8 o’clock is…?  We could write this as 6+8=2 (mod “a clock”).  Basically, Modular Arithmetic is the math of remainders, not quotients.  Thus, as (6+8)/12 has a remainder of 2, 6+8=2 (mod 12).  Similarly, 5*8 = 4 (mod 12), because 40 has a remainder of 4 when divided by twelve. Modular Arithmetic

 Something interesting about working with remainders is how we can work with them. For example, we can evaluate 6*7+8 mod 5 in two ways:  Directly find 6*7+8. Since 6*7+8 = 50, 6*7+8 = 0 (mod 5), because the remainder when 50 is divided by 5 is 0.  Indirectly find 6*7+8 (mod 5). To do this, first notice that 6 = 1 (mod 5), 7 = 2 (mod 5), and 8 = 3 (mod 5). Then, we have 6*7+8 = 1*2+3 (mod 5), and 6*7+8 = 0 (mod 5). This is because the remainder when 1*2+3 = 5 is divided by 5 is 0. Modular Arithmetic

 This works in general because of the distributive property. To see why it works here, write 6 as 5+1, 7 as 5+2, and 8 as 5+3 in our expression 6*7+8, expand, and then evaluate the expression.  6*7 + 8 = (5+1)*(5+2) + (5+3)(mod 5) = (5*5 + 5*2 + 1*5 + 1*2) + (5+3)(mod 5)  Now remove terms with factors of 5 from this sum, because they have a remainder of 0 when divided by 5. We then see that 6*7 + 8 = 1*2+3 mod 5, as desired. Why the Indirect Method Works

 A square is a number we can get if we multiply a number by itself. A square (mod n) is a remainder we can get if we divide a square by n.  So, 4 and 16 are squares no matter what mod we deal with, but some mods have squares that are not normal integer squares – e.g. 2 is a square mod 7, since 3 2 =9, and 9 = 2 (mod 7). Squares in modulo

 Is 2 a square mod 3?  We can notice by checking small examples that 2 never seems to be the remainder of a square. The pattern of remainders of squares mod 3 seems to be 0, 1, 1, 0, 1, 1, 0, …  Think about what we would do if we wanted to evaluate 8 2 mod 3. We can use our indirect method. Since 8 = 2 (mod 3), 8*8 = 2*2 (mod 3). We can use this method no matter how big a number is.  If we are given some number n, we know that n must have a remainder of 0, 1, or 2 when divided by 3. Thus n 2 mod 3 must be one of 0 2, 1 2, or 2 2. Since none of these are equal to 2 mod 3, we know that no square can have a remainder of 2 mod 3.  Thus, 2 is not a square mod 3. Challenge Question

 This is a shorthand way to write a long sum, without ellipses.  So if we were to write the sum of the first n natural numbers, we wouldn’t have to write it as …+n.  i -- This is another way we could write the above sum. Sigma notation Σ n i=1

 Same as sigma notation, but we multiply each term.  Bottom is still the initial value that we iterate, and then we increment and evaluate the sum for each integer until the top integer.  = n! = 1*2*3*…*n Pi Notation Π i n i=1

 We can also write a notation using just the bottom, like this:   Here, the bottom tells us the region in which the integers needed to evaluate this function are, even if that region can’t be defined easily otherwise. Now, the above sum will evaluate to as long as p is 6 or 7. This is because only these integers lie in the range 0≤n≤3 (when p=6) and 0≤n≤3.5 (when p=7). Another Sigma Notation Σ n 0≤n≤p/2

Challenge Question  Write the sum …+m 2 without ellipses, where m=2n-1. n is an arbitrary positive integer.  First, notice that we are summing things together. Thus, we expect to use sigma notation.  Then, each addend is an odd number, squared. The 1 st odd number is 2(1)-1, the 2 nd is 2(2)-1, and so on, so the n th odd number is 2n-1. So, each addend is (2i-1) 2, for each integer i between 1 and n.  The sum is thus Σ (2i-1) 2. i=1 n

 Floor notation is a function. Just like any other function it returns a number based on its input.  The floor function of a number n is the greatest integer that is less than or equal to n, and is usually denoted by | n |.  For example, | 3 | = 3, | 3.9 | = 3, | 0 | = 0, and | -1.4 | = -2. Floor Notation

 Write in standard sigma notation, using the floor function.  The sum starts at 0, so n=0 is at the bottom of the Σ.  The sum ends at the last integer less than or equal to p/2, which is by definition the floor of p/2.  The sum is Challenge question Σ n 0≤n≤p/2 Σ n=0 | p/2 | n.

 Remember, a prime number is an integer whose only positive factors are 1 and itself.  So, 2, 3, and 17 are all primes. But 6 is not, because 6 = 2*3. Since 2 divides 6, we also call 2 a factor of 6. So, 6 has some factors that are not 1 or itself (6), and it is not a prime.  In general, the variables p and q will refer to odd primes from now on, as is custom in number theory. Primes

 Quadratic reciprocity will help us learn when a given number is a square mod an odd prime number. So, it can help us find out if 53 is a square mod 101, for instance.  More importantly, the quadratic formula still holds when we consider a quadratic, mod an odd prime number. So, we can find out how many solutions an equation like x 2 -x-7=0 has in mod 13 (or any odd prime). Why Quadratic Reciprocity?

 Define Legendre symbols to evaluate to 1 if a is a square mod p; otherwise, if a is a nonsquare mod p, this should be -1. In this powerpoint, a Legendre symbol looks like a||p, although usually it looks like a fraction surrounded by parenthesis.  Quadratic Reciprocity states that, if p and q are two distinct odd primes, Legendre Symbols p||q * q||p = (-1) (p-1)(q-1) 4