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Is there a way to determine if there are or aren't any integer positive solutions $(x,y)$ to the equation $a^2x^2 + bx = y^2$ depending on the values of $a$ and $b$?

I tried to deal with it using Pell equations but I just couldn't work it out. Any help is appreciated.

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    $\begingroup$ If you complete the square on the left, you get something like $(ax+c)^2=y^2+d^2$, when $c$ and $d$ depend on $a$ and $b$. So it looks like finding Pythagorean triple. $\endgroup$ – Paul Sep 27 '18 at 14:53
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    $\begingroup$ We have $(2a^2x+b-2ay)(2a^2x+b+2ay)=b^2$. So, every positive integer solution (if any) is of the form $x=\frac{(s-b)^2}{4a^2s},y=\frac{(b+s)(b-s)}{4as}$ where $s$ is a divisor of $b^2$ with $0\lt s\lt |b|$. $\endgroup$ – mathlove Sep 27 '18 at 15:58
  • $\begingroup$ Thanks! I am also wondering about one more similar equation: $a^2x^2+bx=y^2+2^k$ with fixed $k$ (and fixed $a,b$). is it possible to generate a similar conclusion? $\endgroup$ – oren1 Sep 28 '18 at 20:11
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    $\begingroup$ Yes, we get $(2a^2x+b-2ay)(2a^2x+b+2ay)=a^22^{k+2}+b^2$ similarly. $\endgroup$ – mathlove Sep 30 '18 at 15:03
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Above equation shown below has solution:

$a^2x^2 + bx = y^2$

$x=3(k-3)^2$

$y=3(k-3)(3k-4)$

$a=2$

$b=15(k^2-4)$

For $k=5$ we get:

$(x,y,a,b)= (12,66,2,315)$


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  • $\begingroup$ The question is asking how to tell, if we know (a,b), whether there is a corresponding (x,y) to make a solution. $\endgroup$ – Carl Mummert Sep 27 '18 at 20:45
  • $\begingroup$ Yes it's not fully solving my question but it is a help, thanks. can generate something similar for $a^2x^2+bx=y^2+2^k$? $\endgroup$ – oren1 Sep 28 '18 at 20:25
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"OP" has follow up question for equation shown below:

$a^2x^2+bx=y^2+w$

For w=2^6 & w=2^12 there are solutions shown below:

$x=k^2-10k+29$

$y=5k^2-33k+20$

$a=2$

$b=25(k^2-4)$

$w=(2k^2-8)^2$

For k=6 we get w= 2^12 and :

$(x,y,a,b)= (5,2,2,800)$

For k=0 we get w =2^6 and,

$(x,y,a,b)= (29,20,2,-100)$

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