How to find all integer solutions for the equation
$y = \frac{a+bx}{b-x}$, where a and b are known integer values?
P.S. x and y must be integer at the same time
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1$$y=-\frac{b(b-x)}{b-x}+\frac{a+b^2}{b-x}$$ – lone student May 18 '21 at 09:56
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Connected: this question/answers – Jean Marie May 18 '21 at 10:18
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Clear denom's then complete the product as explained in the linked dupes. – Bill Dubuque May 18 '21 at 10:19
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1I don't agree to close this question on the basis that it is the same as the two cited questions which are very different: here, we have a (special) parametric equation... – Jean Marie May 18 '21 at 10:34
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@JeanMarie Exactly the same method(s) apply here, e.g. complete the product, as I said. We already have hundreds of questions showing how to solve such Diophantine equations (e.g. the linked dupes). Nothing is novel here - see abstract duplicates – Bill Dubuque May 19 '21 at 00:01
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First multiply denominator to other side : $0 = yx + bx - by + a = (x-b)(y+b) + a + b^2$
$(x-b)(y+b) = -(a+b^2)$
Then all you need is to write RHS as multiplication of 2 integers: $-(a+b^2) = mn$ and then get 2 solutions $(m+b, n-b)$ and $(n+b, m-b)$ for all different $(m, n)$ pairs.
Corner case: $a = -b^2$, then all $(x,-b)$ is solution except $x=b$ since function is not defined at $x=b$
Snowball
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@JeanMarie we are doing $xy +bx - by - b^2 + a + b^2 = (x-b)(y+b) + a + b^2 = 0$ – Snowball May 18 '21 at 10:26
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@Snowflake with your pseudo, are you completely objective towards Snowball :) ? – Jean Marie May 18 '21 at 14:51
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