Is There Always a Positive Solution for the Linear Diophantine Equation ax+by=c?

  • Thread starter StellaLuna
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In summary, given the conditions that ab<0, d(gcd of a and b)│c, and the equation ax+by=c, there will always be at least one solution with positive values for x and y. This can be achieved by finding a solution to the equation ax-by=1, which can be done using Fermat's Little Theorem and Euler's Totient Function.
  • #1
StellaLuna
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I was wondering if someone could help me with a proof.

If ab<0 (can we assume that either a or b is negative then?) and d(gcd of a and b)│c, there there is at least one solution of ax+by=c with x and y positive.
 
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  • #2


I haven't read it yet but https://www.physicsforums.com/showthread.php?t=367945", is most certainly relevant to your question.

Here is my attempt, which to be honest is a complete overkill since it resorts to a more advanced theorem!:), but it was pretty much the first thing that came into my head.

First we note that if we have one solution to the equation, say (x,y), then (x+nb, y+na) is also a solution. Therefore if we can find one solution, we can always, from this solution, construct another solution for which both coordinates are positive.

With the aim of finding a single solution, we note that it is clear that the problem is equivalent to (where I have redefined the meaning of the variables a,b,c):

[tex]
ax - by = c
[/tex]

For a>0, b>0, and where a and b are coprime [I have divided both sides of the equation by the gcd(a,b), and then redefined the variables (e.g. c now represents the result of this division on the rhs etc..likewise for a,b)].

It is also clear that if we can find a solution to the equation :

[tex]
ax - by = 1
[/tex]

(where a>0, b>0, and where a and b are coprime), then we can immediately find a solution to the equation above (simply by multiplying both sides of the equation by c).

We proceed by noting that from a "[URL of fermat's little theorem[/URL], and given the coprimality of a and b :

[tex]
\exists\ p\ s.t.\ a^p = 1 (mod\ b)
[/tex]

In particular p is given by http://en.wikipedia.org/wiki/Euler%27s_totient_function" : [tex]p=\phi(b)[/tex]. Therefore :

[tex]
a^p = Nb + 1\ for\ some\ integer\ N
[/tex]

So that

[tex]
a^p - N b = 1
[/tex]

Therefore the solution to :

[tex]
ax - by = 1
[/tex]

(where a>0, b>0, and where a and b are coprime) is given by :

[tex]
(x,y) = (a^{p-1}, N)
[/tex]

where x, y are both integers, and therefore we have achieved the desired result
 
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Related to Is There Always a Positive Solution for the Linear Diophantine Equation ax+by=c?

1. What is the definition of "Least one solution of ax+by=c"?

The term "least one solution of ax+by=c" refers to the set of numbers that satisfy the linear equation ax+by=c. In other words, it is the value of x and y that make the equation true when substituted in.

2. How do you find the least one solution of ax+by=c?

To find the least one solution of ax+by=c, you can use the substitution method or the elimination method. In the substitution method, you solve for one variable in terms of the other and then substitute the expression into the other equation. In the elimination method, you add or subtract the two equations to eliminate one variable and then solve for the remaining variable.

3. Can a linear equation have more than one least one solution?

No, a linear equation can only have one least one solution. This is because a linear equation represents a straight line, which can only intersect with another line at one point.

4. What happens when the coefficients a and b in ax+by=c are equal to 0?

If both a and b are equal to 0, the equation becomes 0x+0y=c, which simplifies to 0=c. This means that there is no unique solution to the equation, as any value of x and y would make the equation true.

5. Can a linear equation have no solution?

Yes, a linear equation can have no solution if the lines represented by the equation are parallel and do not intersect. In this case, the lines have different slopes and will never intersect at any point.

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