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How many pairs of positive integers (x,y) satisfy the equation x^2-y^2=18?

 Jun 20, 2021
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Because x and y > 0, x must be greater than y.  

 

We factor x^2-y^2=18 into

 

(x+y)(x-y)=18

 

Because x and y both must be positive integers, x+y and x-y are also both positive integers.  Using this, we also find that x+y >x-y

 

18 factors into 

 

1*18 -> x+y = 18, x-y = 1.  No integer solution

 

2*9 ->  x+y = 9, x-y = 2, No integer solution.  

 

3*6 -> x+y = 6, x-3 = 3, No integer solution.  

 

Thus, we find that there are no solutions.  

 

0 pairs total.  

 Jun 20, 2021

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