Integer solutions to ax^2 + bx - cy^2 - dy = 0

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The discussion focuses on finding integer solutions to the equation ax^2 + bx - cy^2 - dy = 0, specifically where all variables are non-zero integers. A method is proposed that involves analyzing the relationship between the expressions ax^2 + bx and cy^2 + dy, noting a pattern in their differences. The equation can be rearranged to x(ax+b) - y(cy+d) = 0, suggesting a potential use of modular arithmetic to identify common factors between the terms. While the trivial solution (0, 0) exists, the challenge lies in determining non-zero integer values for x and y. The conversation emphasizes the need for a systematic approach to uncover these integer solutions.
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Homework Statement


I am a hobbyist looking for solutions to ax^2 + bx - cy^2 - dy = 0 where all variables are integers and are non-zero. Is there a method of doing this effectively?

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I can look at the numbers produced by ax^2 + bx vs cy^2 + dy and see that they have a relationship: what I mean is if I manually find a pair of close numbers, difference = d, I find the next set of values is d+2 apart, then d + 4 and so on. So it looks as though there should be a method in algebraic terms for doing this.
 
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Of course, the trivial solution is (0, 0). Otherwise, rewrite the equation as x(ax+b)-y(cy+d)=0.
 
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Sorry I know the variables a, b, c, and d but I don't know x, y. It looks as though I might be able to do something with modular arithmetic given that both x(ax+b) and y(cy+d) now seem to both be integer multiples in other words either x or ax+b must necessarily contain some factors in common with y and cy+d. Is there a good way to find x,y? Thanks!
 
http://www4a.wolframalpha.com/Calculate/MSP/MSP100420ag0a9de184i4e300006aa5e486371cg88e?MSPStoreType=image/gif&s=23&w=258.&h=46.
You can find solutions of y in the same way. Its a bit silly though as it requires you to know all but x. You can see intuitively the set of solutions from the form Svein put it in
x(ax+b)−y(cy+d)=0
Yeah you could do what your saying and then write out the set of solutions
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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