Ordered Pairs for x and y: How to Solve x2 + 2x + 18 = y2 with Integers?

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The discussion revolves around finding integer ordered pairs (x, y) that satisfy the equation x² + 2x + 18 = y². A suggestion was made to complete the square, leading to the realization that the equation can be transformed into a difference of squares set equal to -17. The user attempted to find solutions through trial and error, initially proposing pairs (9, 9) and (-9, -7), but these were later confirmed as incorrect. The conversation emphasizes the need for careful algebraic manipulation to identify valid integer solutions. Further clarification on completing the square is sought to ensure accurate results.
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Homework Statement


Please find all ordered pairs of integers (x,y) such that x2 + 2x + 18 = y2

Homework Equations


x2 + 2x + 18 = y2

The Attempt at a Solution



Im not sure on how to begin. I tried square-rooting the expression, to solve for y, turning it into y = \sqrt{x + 2x + 18}. I would greatly appreciate any help to get me started on this problem.

Thanks!
 
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spoc21 said:

Homework Statement


Please find all ordered pairs of integers (x,y) such that x2 + 2x + 18 = y2


Homework Equations


x2 + 2x + 18 = y2


The Attempt at a Solution



Im not sure on how to begin. I tried square-rooting the expression, to solve for y, turning it into y = \sqrt{x + 2x + 18}. I would greatly appreciate any help to get me started on this problem.

Thanks!
It might be helpful to complete the square. Then you would have the difference of two squares being equal to -17.
 
Hey, thanks for the help.
So after this step, do we use trial and error to find the correct solution. I am getting one ordered pair, (9,9), and (-9,-7) after all the math(since x, and y can only be integers). I would really appreciate it if you could confirm this.
Thank you
 
Those ordered pairs you got aren't solutions. When you completed the square, what did you get for your equation?
 

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