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How many pairs of positive integers (x,y) satisfy x^2-y^2=51+xy?

 Jun 14, 2022
 #1
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+1

Hello, 

 

\(x^2-y^2=51+xy\)

 

\(x^2-y^2-xy=51\)

 

\(x^2-yx-y^2-51=0\)

 

Solve with the quadratic formula:

 

\(x_{1,\:2}=\frac{-\left(-y\right)\pm \sqrt{\left(-y\right)^2-4\cdot \:1\cdot \left(-y^2-51\right)}}{2\cdot \:1}\)

 

\(x_{1,\:2}=\frac{-\left(-y\right)\pm \sqrt{5y^2+204}}{2\cdot \:1}\).

 

I think you got \(y\)...

 

Let me know if you need help becoz this might get very messy.Good luck

 Jun 14, 2022
 #2
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This answer is not helpful

Guest Jun 14, 2022
 #3
avatar+556 
+1

How was it not helpful, what else do you need? laugh

Vinculum  Jun 14, 2022

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