Gauss and Germain Raymond Flood Gresham Professor of Geometry Sophie Germain 1st April, 1776 27th June, 1831 Carl Friedrich Gauss 30th April, 1777 23rd February, 1855

Gauss at age 26 Carl Friedrich Gauss 30th April, 1777 23rd February, 1855 Sum the integers from 1 to 100 : Gauss noted they can be arranged in fifty pairs each summing to 101 101 = 1 + 100 = 2 + 99 = =

50 + 51 So the answer is 50 x 101 = 5050 Gauss at age 26 Carl Friedrich Gauss 30th April, 1777 23rd February, 1855 Ceres

Gauss at age 26 Carl Friedrich Gauss 30th April, 1777 23rd February, 1855 Ceres Orbit of Ceres from Gausss notebooks

Disquisitiones Arithmeticae (Discourses in Arithmetic), 1801 Mathematics is the queen of the sciences, and arithmetic the queen of mathematics Modular Arithmetic

We define a b mod n and say a is congruent to b mod n whenever a and b have the same remainders when we divide by n. An equivalent way of describing this is that n divides a b. Modular Arithmetic We define a b mod n and say a is congruent to b mod n whenever a and b have the same remainders when we divide by n. An equivalent way of describing this is

that n divides a b. Examples: 35 11 mod 24, 18 11 mod 7, 16 1 mod 5 Modular Arithmetic We define a b mod n and say a is congruent to b mod n whenever a and b have the same remainders when we divide by n. An equivalent way of describing this is that n divides a b. Examples: 35 11 mod 24, 18 11 mod 7,

16 1 mod 5 If a b mod n and c d mod n Modular Arithmetic We define a b mod n and say a is congruent to b mod n whenever a and b have the same remainders when we divide by n. An equivalent way of describing this is that n divides a b. Examples: 35 11 mod 24, 18 11 mod 7,

16 1 mod 5 If a b mod n and c d mod n Addition: a + c b + d mod n Modular Arithmetic We define a b mod n and say a is congruent to b mod n whenever a and b have the same remainders when we divide by n. An equivalent way of describing this is that n divides a b.

Examples: 35 11 mod 24, 18 11 mod 7, 16 1 mod 5 If a b mod n and c d mod n Addition: a + c b + d mod n Multiplication: ac bd mod n Cancellation or division This is trickier! It is true that 10 4 mod 6 but we cannot divide by 2 to get 5 2 mod 6 as it is not true!

If ac bc mod n then we know that n divides ac bc which is (a b)c but we cannot conclude that n divides (a b) unless n and c have no factors in common. In particular this will happen if n is a prime and n does not divide into c. Modular Arithmetic We define a b mod n and say a is congruent to b mod n whenever a and b have the same remainders when we divide

by n. An equivalent way of describing this is that n divides a b. Examples: 35 11 mod 24, 18 11 mod 7, 16 1 mod 5 If a b mod n and c d mod n Addition: a + c b + d mod n Multiplication: ac bd mod n Quadratic residues 11 12 mod 5

so we can think of 1 as being the square root of 11 in mod 5 arithmetic 7 62 mod 29 so we can think of 6 as being the square root of 7 in mod 29 arithmetic Quadratic residues 11 12 mod 5 so we can think of 1 as being the square root of 11 in mod 5 arithmetic

7 62 mod 29 so we can think of 6 as being the square root of 7 in mod 29 arithmetic Definition: p is a quadratic residue of q if p is congruent to a square modulo q i.e. there is an integer x so that p x2 mod q 11 is a quadratic residue modulo 5 7 is a quadratic residue modulo 29 5 42 mod 11 so 5 is a quadratic residue modulo 11 Quadratic Reciprocity Theorem The quadratic reciprocity theorem is

concerned with when two primes p and q have a square root modulo each other Example: 13 and 29 have a square root modulo each other since 29 16 mod 13 and 13 100 mod 29 Two families Primes congruent to 1 mod 4 5 13 17 29 37 41 53 61 73 79 89 97 101 Primes congruent to 3 mod 4

3 7 11 19 23 31 43 47 59 67 71 83 103 Primes congruent to 1 mod 4 5 13 17 29 37 41 53 61 73 79 89 97 101 Two primes from this family either both have a square root modulo the other or neither has: If the primes are p and q then p x2 mod q has a solution if and only if q

x2 mod p does Primes congruent to 1 mod 4 5 13 17 29 37 41 53 61 73 79 89 97 101 Two primes from this family either both have a square root modulo the other or neither has: If the primes are p and q then p x2 mod q has a solution if and only if q x2 mod p does

Example: For 13 and 29 they both have as 29 16 mod 13 and 13 100 mod 29 Primes congruent to 1 mod 4 5 13 17 29 37 41 53 61 73 79 89 97 101 Two primes from this family either both have a square root modulo the other or neither has: If the primes are p and q then p x2 mod q has a solution if and only if q x2 mod p does

Example: For 5 and 13 neither of them has just write out the square residues i.e. calculate all the residues x2 mod 5 and x2 mod 13 Primes congruent to 3 mod 4 3 7 11 19 23 31 43 47 59 67 71 83 103 For two primes from this family one

and only one of them has square root modulo the other. If the primes are p and q then p is a square mod q if and only if q is not a square mod p. Primes congruent to 3 mod 4 3 7 11 19 23 31 43 47 59 67 71 83 103 For two primes from this family one and

only one of them has square root modulo the other. If the primes are p and q then p is a square mod q if and only if q is not a square mod p. Example: p = 7 and q = 11. Now 11 22 mod 7 but there is no x so that 7 x2 mod 11. Just calculate for every x! One prime from each family Primes congruent to 1 mod 4 5 13 17 29 37 41 53 61 73 79 89 97 101

Primes congruent to 3 mod 4 3 7 11 19 23 31 43 47 59 67 71 83 103 The situation is the same as if they were both from the first family. For 29 and 7 they both have. 7 62 mod 29 and 29 12 mod 7 Primes in Arithmetic Progressions Dirichlet in 1837 made essential use of the theorem of quadratic reciprocity to prove that any arithmetic progression

a, a + d, a + 2d, a + 3d, , where a and d have no common factors contains infinitely many primes Richard Taylor developed with Andrew Wiles the TaylorWiles method, which they used to help complete the proof of Fermats Last Theorem. Battle of Jena, 1806 Duke of Brunswick 1735 - 1806

Gauss letter of 30 April, 1807 The scientific notes with which your letters are so richly filled have given me a thousand pleasures. I have studied them with attention and I admire the ease with which you penetrate all branches of arithmetic, and the wisdom with which you generalize

and perfect. Fermats marginal note Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem

in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general,

any power higher than the second, into two like powers. I have discovered a truly marvellous proof of this, which this margin is too narrow to contain. Fermats Last Theorem The equation

xn + yn = zn has no whole number solutions if n is any integer greater than or equal to 3. Only need to prove for n equal to 4 or n an odd prime. Suppose you have proved the theorem when n is 4 or an odd prime then it must also be true for every other n for example for n = 200 because x200 + y200 = z200 can be rewritten (x50)4 + Let p be an odd prime such that 2p + 1 is also

prime. Then xp + yp = zp implies p divides xyz Here is what I have found: The Plan For any given prime, p, to find an infinite number of other primes, called auxiliary primes, satisfying particular conditions These conditions would allow her to deduce that

each of these auxiliary primes would have to divide one of x, y or z in any solution But since there are an infinite number of them there can be no solution! If x3 + y3 = z3 then 7 divides xyz 7 satisfies the condition to be an auxiliary prime for 3

The structure of the proof is to assume that 7 does not divide xyz and obtain a contradiction, forcing us to conclude that 7 does divide xyz x3 + y3 z3 mod 7 Now a3 1 mod 7 whenever 7 does not divide a 13 1 mod 7, 23 mod 7, 33 mod 7, 43 mod 7, 53 mod 7,and 63 mod 7 So x3 + y3 z3 mod 7 becomes + mod 7 which is impossible so 7 must divide xyz

Sophie Germain was a much more impressive number theorist than anyone has ever previously known. Chladni patterns from his 1809 book Trait dAcoustique

Formulate a mathematical theory of elastic surfaces and indicate just how it agrees with empirical evidence Sophie Germain's sketch of an elastic bar and its radius of curvature when bent by an external force, taken from her book on the theory of elastic

Curvature at a point on a curve Define the curvature at each point of a circle to be 1 over the radius Given a curve, C, and a point, P, on it there is a unique circle or line which

most closely approximates the curve near P. Define the curvature at P to be the curvature of that Curvature on a surface Different normal planes will give different curves and out of all these different curves one will have greatest curvature and one will have least curvature. They are called the principal directions and the principal curvatures at the point are the

curvatures in these directions. Curvature on surfaces Cylinder: Principal curvatures are 0 and 1 over radius of cylinder Sphere: Principal

curvatures are both 1 over radius of sphere Saddle: Principal curvatures are 1/r and 1/r' If the two circles are on the opposite side of the surface, we say the curvature is

The efforts of Sophie Germain, of Navier and of Poisson were part of a historical process that involved rigourous analysis, competent experiment, ingenious hypotheses, fertile concepts, and the extension of mathematical prowess: a process, also, that involved

intellectual prejudice, personal antipathy, political maneuvering, and the use of position for the benefit of friends. Out of all this a theory of elasticity emerged. Sophie Germain Aged 14 , Illustration from "Histoire Du Socialisme," circa 1880, by Auguste Eugene Leray

A statue of Sophie Germain standing in the courtyard of the Ecole Sophie Germain, a lyce in Paris 1 pm on Tuesdays at the Museum of London Einsteins Annus Mirabilis, 1905 Tuesday 20 October 2015 Hamilton, Boole and their Algebras

Tuesday 17 November 2015 Babbage and Lovelace Tuesday 19 January 2016 Gauss and Germain Tuesday 16 February 2016 Hardy, Littlewood and Ramanujan Tuesday 15 March 2016 Turing and von Neumann Tuesday 19 April 2016