DIOPHANTINE EQUATIONS KUCERA PDF
the role of diophantine equations in the synthesis of feedback control systems. 12 20 18 atom c. e-mail [email protected] that evolve in discrete time. This relationship, termed canonical Diophantine equations, can be used to resolve a  V. KUCERA, Discrete Linear Control, John Wiley,New York, of linear control systems has revied an interest in linear Diophantine equations for polynomials. Vladimir Kučera; Jan Ježek; Miloš Krupička.
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Solving a linear Diophantine equation means that you need to find solutions for the variables x and y that are integers only. Finding integral solutions is more difficult than a standard solution and requires an ordered pattern of steps.
You must first equatlons the greatest common factor of the coefficients in the problem, and then use that result to find a solution. If you can find one integral solution to a linear equation, you can apply a simple pattern to find infinitely many more.
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Together, they cited information from 17 referenceswhich can be found at the bottom of the page. Write the equation in standard form.
A linear equation is one that has no exponents greater than 1 on any variables. If the equation is not already in standard form, you need to use the basic rules of algebra to rearrange or combine the terms to create the standard form.
Reduce the equation if possible. If there is a common factor in all three terms, then reduce the equation by dividing all terms by that factor. If you reduce evenly across all three terms, then any solution you find for the reduced equation will also be a solution for the original equation. For example, if all three terms are even, you can at least divide by 2, as follows: Check for the impossibility of a solution. In some cases, you may be able to tell immediately if there is no solution to your problem.
If you see a common factor on the left side of the equation that is not shared on the right side, then there can be no solution to the problem. Review the Euclidean algorithm. The Euclidean algorithm is a system of repeated divisions, using the remainder each time as the divisor of a new division. The last divisor that divides evenly is the greatest common factor GCF of the two numbers. Note the new remainder Divide the previous divisor 20 by the previous remainder Note the new remainder 4.
Divide the previous divisor 16 by the previous remainder 4. Since the remainder is now 0, conclude that 4 is the GCF of the original two numbers and Apply the Euclidean algorithm to the coefficients A and B.
With your linear equation in standard form, identify the coefficients A and B.
Diophantine equations in control – A survey
Apply the Euclidean algorithm to find their GCF. Identify the greatest common factor GCF. Because the Euclidean algorithm for this pair continues all the way down to dividing by 1, the GCF between 87 and 64 is 1.
This is another way of saying that 87 and 64 are relatively prime. If not, then there will be no solution. Suppose, for example, that the GCF had worked out to be 5. The divisor 5 cannot go evenly into 3. In that case, the equation would have no integral solutions. As you will see below, if an equation has one integral solution, then it also has infinitely many integral solutions. Label the steps of the GCF reduction.
To find the solution of the linear equation, you will use your work on the Euclidean algorithm as the basis for a repeated process of renaming and simplifying values. Thus, you have the following steps: Begin with the last step that has a remainder.
Rewrite that equation so the remainder stands alone, as equal to the rest of the information in the equation.
That remainder was 1. Rewrite the equation in Step 6 as follows: Isolate the remainder of the previous step. Each time, you will be revising the right side of the equation in terms of the numbers in the higher step. Perform a substitution and simplify. You should notice that your revision of Step 6 contains the number 2, and your revision of Step 5 is equal to 2. Substitute the equality in Step 5 into the place of the 2 in your Step 6 revision: This is the Step 6 revision.
Substitute in place of the value 2. Repeat the process of substitution and simplification. Moving through the Euclidean algorithm steps in reverse, repeat the process. Each time, you will revise the previous step, and substitute its value into your latest result.
Now revise Step 4 to isolate its remainder as: Continue repeating substitution and simplification. This process will repeat, step by step, until you reach the original step of the Euclidean algorithm.
The purpose of this procedure is to wind up with an equation that will be written in terms of 87 and 64, which are the original coefficients of the problem you are trying to solve. Continuing in this manner, the remaining steps are as follows: Rewrite the result in terms of the original coefficients. When you return to the first step of the Euclidean algorithm, you should notice that the resulting equation contains the two coefficients of the original problem.
Rearrange the numbers so they align with the original equation. Thus, you can rearrange your last step to put the terms in that standard order. Pay particular eqjations to the 64 term. In the original problem, that term is subtracted, but the Euclidean algorithm treats it as a positive term. To account for the subtraction, you need to change the multiplier 34 to a negative.
The final equation looks like this: Multiply by the necessary factor to find your solutions. Notice that the greatest common divisor for this problem was 1, so the solution that you reached is equal to 1. However, that is not the solution to the problem, since diophxntine original problem equatins 87xy equal to 3. You need to multiply the terms of your last equation by diophanine to get a solution: Identify the integral solution to the equation. The values that must be multiplied by the coefficients are the x and y solutions to the equation.
Recognize that infinitely many solutions exist. If a linear equation has one integral solution, then it must have infinitely many integral solutions.
Here is a brief algebraic statement of the proof: Adding a B to x while diophanitne A from y results in the same solution. Identify your original solution values for x and y.
Diophantine equations in control – A survey – Semantic Scholar
The pattern of infinite solutions begins with the single solution that you identified. Add the y-coefficient B to the x solution. To find a new solution for x, add the value of the coefficient of y. Subtract the x-coefficient A from the y solution. To make the equation remain balanced, when you add to the x term, you must fquations subtract from the y term. To verify that your new ordered pair is a solution to the equation, insert the values into the equation and see if it works.
Write a general kuxera. The values for x will fit a pattern of the original solution, plus any multiple of the B coefficient. You can write this algebraically as follows: For this problem, you can say: Introduce a second variable to convert the modular equation to an equivalent diophantine equarion. But 2x – 3y is an integer.
Ewuations left side is always a multiple of 14, but 38 is not. So that equation has no solutions mod Not Helpful 0 Helpful 3.
Both ordinary and diophantine equations can have any type of integer or non-integer coefficients. Diophantine-ness refers to the domain of the variable s – it’s those that have to be integers. Not Helpful 0 Helpful 0. How do I find solutions to word problems involving linear Diophantine equations?
Figure out what the question is asking. Cross out any irrelevant information, then put all the values into your equation. Not Helpful 2 Helpful 0.