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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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
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add a comment |
$begingroup$
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?
diophantine-equations
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That's the same as $(x+c)(y+b)=bc-d$, so really it is just a factoring problem.
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– Lord Shark the Unknown
Apr 2 at 15:56
add a comment |
$begingroup$
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?
diophantine-equations
$endgroup$
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
diophantine-equations
asked Apr 2 at 15:06
Quote DaveQuote Dave
336
336
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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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint
Assuming $b,c,d$ are given integers,
$$(b+y)(c+x)=bc-d$$
Or
$$y=-dfracbx+dc+x=-b+?$$
$endgroup$
$begingroup$
I don't understand what you mean by this. Could you clarify?
$endgroup$
– Quote Dave
Apr 2 at 21:03
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
Assuming $b,c,d$ are given integers,
$$(b+y)(c+x)=bc-d$$
Or
$$y=-dfracbx+dc+x=-b+?$$
$endgroup$
$begingroup$
I don't understand what you mean by this. Could you clarify?
$endgroup$
– Quote Dave
Apr 2 at 21:03
add a comment |
$begingroup$
Hint
Assuming $b,c,d$ are given integers,
$$(b+y)(c+x)=bc-d$$
Or
$$y=-dfracbx+dc+x=-b+?$$
$endgroup$
$begingroup$
I don't understand what you mean by this. Could you clarify?
$endgroup$
– Quote Dave
Apr 2 at 21:03
add a comment |
$begingroup$
Hint
Assuming $b,c,d$ are given integers,
$$(b+y)(c+x)=bc-d$$
Or
$$y=-dfracbx+dc+x=-b+?$$
$endgroup$
Hint
Assuming $b,c,d$ are given integers,
$$(b+y)(c+x)=bc-d$$
Or
$$y=-dfracbx+dc+x=-b+?$$
answered Apr 2 at 15:08
lab bhattacharjeelab bhattacharjee
229k15159279
229k15159279
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I don't understand what you mean by this. Could you clarify?
$endgroup$
– Quote Dave
Apr 2 at 21:03
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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