Diophantine Equations of Degree 2 Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Strange Cubic Diophantine EquationsSecond degree Diophantine equationsDiophantine equation without unique formula for solutionsDiophantine Equations involving cubesHow to solve simultaneous modulus / diophantine equationsClassifying Diophantine EquationsFinding other solutions to diophantine equationsHyperbolic Diophantine Equations: Application of Euclidean Algorithm?Solution to rational Diophantine equations in fixed pointInteger solutions to a Diophantine Equation

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Diophantine Equations of Degree 2



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Strange Cubic Diophantine EquationsSecond degree Diophantine equationsDiophantine equation without unique formula for solutionsDiophantine Equations involving cubesHow to solve simultaneous modulus / diophantine equationsClassifying Diophantine EquationsFinding other solutions to diophantine equationsHyperbolic Diophantine Equations: Application of Euclidean Algorithm?Solution to rational Diophantine equations in fixed pointInteger solutions to a Diophantine Equation










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$begingroup$


During my studies, I have seen equations of this form $$xy+bx+cy+d=0$$ Is there a way to find solutions of equations of this form or the number of solutions of equations of this form without factoring or checking within a range?










share|cite|improve this question









$endgroup$











  • $begingroup$
    That's the same as $(x+c)(y+b)=bc-d$, so really it is just a factoring problem.
    $endgroup$
    – Lord Shark the Unknown
    Apr 2 at 15:56















0












$begingroup$


During my studies, I have seen equations of this form $$xy+bx+cy+d=0$$ Is there a way to find solutions of equations of this form or the number of solutions of equations of this form without factoring or checking within a range?










share|cite|improve this question









$endgroup$











  • $begingroup$
    That's the same as $(x+c)(y+b)=bc-d$, so really it is just a factoring problem.
    $endgroup$
    – Lord Shark the Unknown
    Apr 2 at 15:56













0












0








0


1



$begingroup$


During my studies, I have seen equations of this form $$xy+bx+cy+d=0$$ Is there a way to find solutions of equations of this form or the number of solutions of equations of this form without factoring or checking within a range?










share|cite|improve this question









$endgroup$




During my studies, I have seen equations of this form $$xy+bx+cy+d=0$$ Is there a way to find solutions of equations of this form or the number of solutions of equations of this form without factoring or checking within a range?







diophantine-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 2 at 15:06









Quote DaveQuote Dave

336




336











  • $begingroup$
    That's the same as $(x+c)(y+b)=bc-d$, so really it is just a factoring problem.
    $endgroup$
    – Lord Shark the Unknown
    Apr 2 at 15:56
















  • $begingroup$
    That's the same as $(x+c)(y+b)=bc-d$, so really it is just a factoring problem.
    $endgroup$
    – Lord Shark the Unknown
    Apr 2 at 15:56















$begingroup$
That's the same as $(x+c)(y+b)=bc-d$, so really it is just a factoring problem.
$endgroup$
– Lord Shark the Unknown
Apr 2 at 15:56




$begingroup$
That's the same as $(x+c)(y+b)=bc-d$, so really it is just a factoring problem.
$endgroup$
– Lord Shark the Unknown
Apr 2 at 15:56










1 Answer
1






active

oldest

votes


















0












$begingroup$

Hint



Assuming $b,c,d$ are given integers,



$$(b+y)(c+x)=bc-d$$



Or
$$y=-dfracbx+dc+x=-b+?$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    I don't understand what you mean by this. Could you clarify?
    $endgroup$
    – Quote Dave
    Apr 2 at 21:03











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Hint



Assuming $b,c,d$ are given integers,



$$(b+y)(c+x)=bc-d$$



Or
$$y=-dfracbx+dc+x=-b+?$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    I don't understand what you mean by this. Could you clarify?
    $endgroup$
    – Quote Dave
    Apr 2 at 21:03















0












$begingroup$

Hint



Assuming $b,c,d$ are given integers,



$$(b+y)(c+x)=bc-d$$



Or
$$y=-dfracbx+dc+x=-b+?$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    I don't understand what you mean by this. Could you clarify?
    $endgroup$
    – Quote Dave
    Apr 2 at 21:03













0












0








0





$begingroup$

Hint



Assuming $b,c,d$ are given integers,



$$(b+y)(c+x)=bc-d$$



Or
$$y=-dfracbx+dc+x=-b+?$$






share|cite|improve this answer









$endgroup$



Hint



Assuming $b,c,d$ are given integers,



$$(b+y)(c+x)=bc-d$$



Or
$$y=-dfracbx+dc+x=-b+?$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 2 at 15:08









lab bhattacharjeelab bhattacharjee

229k15159279




229k15159279











  • $begingroup$
    I don't understand what you mean by this. Could you clarify?
    $endgroup$
    – Quote Dave
    Apr 2 at 21:03
















  • $begingroup$
    I don't understand what you mean by this. Could you clarify?
    $endgroup$
    – Quote Dave
    Apr 2 at 21:03















$begingroup$
I don't understand what you mean by this. Could you clarify?
$endgroup$
– Quote Dave
Apr 2 at 21:03




$begingroup$
I don't understand what you mean by this. Could you clarify?
$endgroup$
– Quote Dave
Apr 2 at 21:03

















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