CSE 311: Foundations of Computing I Solving Modular PDF

Solving Modular Equivalences. Solving a Normal Equation. First we discuss an analogous type of question when using normal arithmetic. Question: Solve the

MATH 565 Spring 2019 - Class Notes Solving Linear Equations

13 mars 2019 Summary: This class covered how to solve linear equations modulo n ... We can not divide by a in modular arithmetic so how can we cancel out ...

SOLVING SIMULTANEOUS MODULAR EQUATIONS OF LOW

SOLVING SIMULTANEOUS MODULAR EQUATIONS OF LOW DEGREE. Johan Hastad*. MIT. Abstract: We consider the problem of solving systems of equations Pi(x).

modular-arithmetic.pdf

Solve the congruence. 6x +1=2(x + 2) (mod 7) . The modular arithmetic properties allow me to solve this equation the way I would solve a linear equation up to

CSE 311 Lecture 14: Euclidean Algorithm and Modular Equations

A quick review of . Extended Euclidean algorithm. Bézout's theorem and the extended Euclidean algorithm. Modular equations. Solving modular equations with the

Finding a Small Root of a Univariate Modular Equation

We show how to solve a polynomial equation (mod N ) of degree k in a single variable z as long as there is a solution smaller.

Finding a small root of a univariate modular equation

We show how to solve a polynomial equation (mod N) of degree k in a single variable x as long as there is a solution smaller than N¹/*.

Small Solutions to Polynomial Equations and Low Exponent RSA

In the bivariate integer case we combine c(x0 y0) with p(x0

Derivation of modular anomaly equation in compact elliptic Calabi

Combined with the B-model method of holomorphic anomaly equa- tion and boundary conditions we can solve topological strings to very.

Solving Modular Problems

Solving Modular. Problems. Chapter 10 – Section 3. Simple Modular Equations. ? The solution to a modular equation is a set.

[PDF] Modular Equivalences Solving a Normal Equation - Washington

First we discuss an analogous type of question when using normal arithmetic Question: Solve the equation 27y = 12 Solution: We divide both sides by 27 to get

[PDF] Modular Arithmetic

Solve the congruence 6x +1=2(x + 2) (mod 7) The modular arithmetic properties allow me to solve this equation the way I would solve a linear equation up to

[PDF] Computing the modular equation

The modular equation Over C elliptic curves E1 and E2 are related by a cyclic isogeny of degree N if and only if ?N(j(E1)j(E2)) = 0

[PDF] Class Notes Solving Linear Equations Modulo n - TigerWeb

13 mar 2019 · Summary: This class covered how to solve linear equations modulo n using inverses and how to solve systems of concurrences with the Chinese

[PDF] Solving Modular Problems

Solving Modular Problems 1 Solving Modular Problems Chapter 10 – Section 3 Simple Modular Equations ? The solution to a modular equation is a set

(PDF) Solving Systems of Modular Equations in One Variable

PDF We address the problem of polynomial time solving univari- ate modular equations with mutually co-prime moduli For a given sys- tem of equations

[PDF] Examples of Modular Arithmetic

First of all we recall how to solve linear Diophantine equations: Claim 0 (Solving Linear Diophantine Equations in two Variables) Let a and b integers not

[PDF] Explicit formulas for the modular equation - UC Davis Mathematics

EXPLICIT FORMULAS FOR THE MODULAR EQUATION PAUL BAGINSKI AND ELENA FUCHS Abstract We determine an algorithm for calculating the modular equation

[PDF] Solving Systems of Linear Equalities in Modular Arithmetic with

In the first step we start to divide the solution space into p small cells at each added mod p constraint according to modular arithmetic p Then by

[PDF] Ramanujan Modular Equations and Approximations to Pi or How to

This extends in the obvious way to the solution of any algebraic equation with the startling conclusion that every algebraic number can be computed (to n-digit

How to solve Diophantine equations using modular arithmetic?
A mod B = ( A + K ? B ) mod B A \\text{ mod } B = (A + K \\cdot B)

CSE 311: Foundations of Computing I

Solving Modular Equivalences

Solving a Normal Equation

First, we discuss an analogous type of question when using normal arithmetic.

Question:Solve the equation27y= 12.

Solution:We divide both sides by 27 to gety=1227

Solution:We multiply both sides by 1/27 to gety=1227 These solutions are two ways of saying the same thing.

Solving a Modular Congruence

Now, we consider a congruence instead:

Question:Solve the congruence27y10 (mod 4).

Note:We can"t just divide both sides. For example, consider510 (mod 5). If we were to divide both sides by 5, we would get12 (mod 5)which is definitely false. Another way of looking at this would be to ask the question What is 15 mod5? It really doesn"t make any sense, because remainders should always be integers. So, instead, we need to create machinery to multiply by whatever thecorrectinverse is mod a number.

Inverses

Ifxy= 1, we say thatyis the "multiplicative inverse ofx". We have a similar idea modm: Ifxy1 (modm), we sayyis the "multiplicative inverse ofxmodulo m". How do we compute the multiplicative inverse ofxmodulom? By definition,xy1 (modm)iffxy+tm= 1for somet2Z. We know by Bezout"s Theorem that we can findyandtsuch thatxy+tm= gcd(x;m). Said another way: Ifgcd(x;m) = 1, then we can find a multiplicative inverse!

To actually compute the multiplicative inverse, we use the Extended Euclidean Algorithm. For example,

consider the equation we were trying to solve above:27y10 (mod 4). First, we find the multiplicative inverse of 27 modulo 4. That is, we find aysuch that27y1 (mod 4). To do this, we first note that thegcd(27;4) = gcd(4;3) = gcd(3;1) = gcd(1;0) = 1, which means an inverse does exist!

Now, we write out the equations:

27 = 64 + 3

4 = 13 + 1

Solving each equation for the remainder:

3 = 2764

1 = 413

Backward substituting, we get:

1 = 413

= 41(2764) = 74 + (1)27 So, we have found that1mod4 = 3mod4is the multiplicative inverse of 27 modulo 4. We can verify this by taking(273)mod4 = 81mod4 = 1.

Solving the original equation

Now, we need to solve the original equation:27y10 (mod 4). We know from above that2731 (mod 4). So, multiplying both sides by 10 (which works, because of a theorem from lecture; note that this is different than the theorem from the homework!), we get:

273010 (mod 4)

Since30mod4 = 2, we have27210 (mod 4). It follows thatx= 2solves the original equation.

[PDF] CSE 311: Foundations of Computing I Solving Modular

How to solve Diophantine equations using modular arithmetic?

CSE 311: Foundations of Computing I

Solving Modular Equivalences

Solving a Normal Equation

Question:Solve the equation27y= 12.

Solution:We divide both sides by 27 to gety=1227

Solving a Modular Congruence

Now, we consider a congruence instead:

Question:Solve the congruence27y10 (mod 4).

Inverses

Now, we write out the equations:

27 = 64 + 3

4 = 13 + 1

Solving each equation for the remainder:

3 = 2764

1 = 413

Backward substituting, we get:

1 = 413

Solving the original equation

273010 (mod 4)

Other Solutions?