Solve the linear Diophantine equation

  • Thread starter chwala
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That's the same thing you have.In summary, the conversation discusses solving a diophantine equation with multiple variables. The equation is simplified and divided by a common factor to find a solution. The solution is then substituted back into the original equation to check for accuracy.
  • #1
chwala
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Homework Statement
solve the linear diophantine equation ##259x+581y = 7##
Relevant Equations
euclidean algorithm
##259x+581y=7##
##581=259.2+63##
##259=63.4+7##
##63=7.9+0##
therefore by reversing...
##7=0.63+1.7##
##7=1.259-4.63###
##x=1## ##y=-4##
is this correct? Bingo
 
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  • #2
chwala said:
##x=1## ##y=-4##
is this correct? Bingo
What happens when you substitute the solution back into the equation?
 
  • #3
Also, never use a . to indicate multiplication. It makes things very hard to read.
 
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  • #4
i think i figured it out...
##7=1×259-4×63##
##7=1×259-4(581-259×2)##
##7=259×9+581×-4##
on using the general form... we have
##x= 9+83m##
##y = -4-37m##
bingo bingo from Africa:smile:
 
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  • #5
You may have saved some work if you divided through the first equation by 7. Notice condition for solution apply/exist.
 
  • #6
WWGD said:
You may have saved some work if you divided through the first equation by 7. Notice condition for solution apply/exist.
can you show how? i have already posted the solution therefore an alternative method would be appreciated...
 
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  • #7
alternatively from my research you can have
##259x+581y=7##
gcd##(581,259)=7##
therefore dividing the diophantine equation by 7 we have
##83y+37x=1##
83/37= 2+ 1/{37/9}
= 2+ 1/{4+{1/9}}
= 2+ 1/4= 9/4 now assigning suitable change sign on the numerator and denominator we have
##83(-4)+37(9)=1##
 
  • #8
chwala said:
can you show how? i have already posted the solution therefore an alternative method would be appreciated...
Sure, I was referring to the needed conditions for an integer solution for ## ax+by=c ## to exist. We need gcd(a,b)|c . Just for completeness.
 
  • #9
There are few key ideas you need to know for linear diophantine equations:
(1)The Linear Diophantine Equation
aX+bY=1 has a solution iff gcd(a,b)=1.

(2) The Linear Diophantine Equation aX+bY=n has solution iff gcd(a,b)|n.

(3) Euclid's Lemma: Let D be gcd(a,b). Then the Linear Diophantine Equation aX+bY=D has a solution.

There are two more important results. Can you fill them in?
 
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  • #10
MidgetDwarf said:
There are few key ideas you need to know for linear diophantine equations:
(1)The Linear Diophantine Equation
aX+bY=1 has a solution iff gcd(a,b)=1.

(2) The Linear Diophantine Equation aX+bY=n has solution iff gcd(a,b)|n.

(3) Euclid's Lemma: Let D be gcd(a,b). Then the Linear Diophantine Equation aX+bY=D has a solution.

There are two more important results. Can you fill them in?
Thanks for your insight, i should be fine with the idea of solving the diophantine equations...
 
  • #11
Solving the Diophant Equation with Multiple Variables
https://abakbot.com/2019-03-12-19-33-47/diofn
Example
Original diophant equation
mathtex.png
A private solution to this equation
mathtex.png
 
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  • #12
DimsVS said:
Solving the Diophant Equation with Multiple Variables
https://abakbot.com/2019-03-12-19-33-47/diofn
Example
Original diophant equation
A private solution to this equation
how did you get your ##x## values?
 
  • #14
Did you notice that all three of 259, 581, and 7 are multiples of 7? The first thing I would do is divide by 7 to simplify to 37x+ 83y= 1. What I would now do is essentially what you did. 37 divides into 83 twice with remainder 9: 83- 2(37)= 9. 9 divides into 37 four times with remainder 1: 37- 4(9)= 1. Replacing "9" in that by "83- 2(37)" gives 37- 4(83- 2(37))= 9(37)- 4(83)= 1. Comparing that to "37x+ 83y= 1" we have x= 9 and y= -4. Of course then x= 9+ 83n and y= -4- 37n gives 37(9+ 83n)+ 83(-4- 37n)= 37(9)+ (37)(83n)- 83(4)- (37)(83n)= 1.
 

1. What is a linear Diophantine equation?

A linear Diophantine equation is an equation in two or more variables where the variables must be integers and the coefficients of the variables are also integers. The equation can be written in the form ax + by = c, where a, b, and c are integers.

2. How do you solve a linear Diophantine equation?

To solve a linear Diophantine equation, you can use the extended Euclidean algorithm or the method of back substitution. The extended Euclidean algorithm is more efficient and can be used for any type of linear Diophantine equation, while the method of back substitution is more suitable for simpler equations.

3. Can all linear Diophantine equations be solved?

No, not all linear Diophantine equations have solutions. If the coefficients of the variables have a common factor that does not divide the constant term, then the equation has no solutions. For example, the equation 2x + 3y = 7 has no solutions because 7 is not divisible by 2 or 3.

4. What is the importance of solving linear Diophantine equations?

Linear Diophantine equations have various applications in number theory, cryptography, and computer science. They are also useful in solving real-world problems, such as finding the number of solutions to a given set of linear equations.

5. Are there any techniques to make solving linear Diophantine equations easier?

Yes, there are some techniques that can make solving linear Diophantine equations easier. These include using the substitution method, Gaussian elimination, and modular arithmetic. It is also helpful to have a good understanding of number theory and algebraic concepts.

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