Positive solution for linear Diophantine equations

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The discussion focuses on finding positive integer solutions for the linear Diophantine equation ax + by = c, where a, b, and c are natural numbers. It is established that if c is a multiple of gcd(a, b), there are infinite integer solutions, but positive solutions may not always exist. An example provided is the equation 7x + 6y = 5, which has integer solutions but no positive solutions. The method to derive solutions involves expressing them in the form x = x0 + kb and y = y0 - ka, where k is an integer, and the feasibility of positive solutions depends on the values of a, b, x0, and y0. The conversation highlights the complexity of ensuring both x and y remain positive in such equations.
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The linear Diophantine equations: ax+by=c, a,b,c is natural numbers.
If c is a multiple of gcd(a,b), there is infinite integer solutions, and I know how to find x,y.
However, I wonder how to find positive integer solution x,y only.
 
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pyfgcr said:
The linear Diophantine equations: ax+by=c, a,b,c is natural numbers.
If c is a multiple of gcd(a,b), there is infinite integer solutions, and I know how to find x,y.
However, I wonder how to find positive integer solution x,y only.


They may not exists. For example, the equation \,7x+6y=5\, cannot have positive solutions, but it has

solutions, like \,(5,-6)\,

DonAntonio
 
Shouldn't that be (5,-5)?
 
Mensanator said:
Shouldn't that be (5,-5)?


Yes, you're right of course. Thanks.

DonAntonio
 
All solutions of the Diophantine equation ax+ by= c (assuming a, b, relatively prime) are of the form x= x0+ kb, y= y0- ka for k any integer. If you want both x and y positive, you must be able to choose k so that those are postive. Whether that is possible, of course, depends on a, b, x0, and y0.
 
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