Algebra graphing problem ```Name: yendor Status: N/A Age: N/A Location: N/A Country: N/A Date: Around 1995 ``` Question: Could someone give me the necessary and sufficient conditions on the non- zero integers a,b,c such that the graph of ax+by=c contains a point (a',b') where a' and b' are integers. Please give an example or two to demonstrate this. Replies: But if a' and b' are integers, the c is necessarily an integer. On the other hand, if c is an integer, then there are obviously rational (non-inte- ger) values of a' and b' for which c is not an integer. I think that the proposition, as you have stated it, is false. jluI cannot say I understand jlu's point. Anyway, suppose a = 2, b = 4, and c = 3. Then there cannot be any integer solutions for a', b', because ax + by is always even, if x and y are integers, and so can never be equal to 3. I believe the necessary and sufficient conditions are that any common divisor of a and b (such as 2 in the above example) must also divide c evenly. If that is true, then the common divisors can be factored out of the equation, and so one can simply start by assuming a and b have no common divisors - and there is a well-known result that for any two numbers which have this property, you can find integers x and y such that ax + by = 1. Then just multiply both x and y by c and you have the solution you want. So this condition is sufficient. It is obviously necessary, for the same reason as in the example above. asmith Click here to return to the Mathematics Archives

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