Find the ordered pairs

**spoc21** · Sep 18th 2010, 11:45 AM

Hello, I'm having some trouble with the following question, and have not been able to solve it despite excessive attempts:

Find all ordered pairs of integers (x,y) such that $x^2+2x+18=y^2$

I tried square-rooting both sides so that the equation could be solved for y, but didn't have any luck. I also thought about using the "completing the square" method, but am not sure on how to use it in this case. I would greatly appreciate any help.

Thanks!

**Plato** · Sep 18th 2010, 12:00 PM

I found one pair. There may be others.
Look at $(x+1)^2=y^2-17$ .

**HallsofIvy** · Sep 18th 2010, 01:26 PM

Plato's suggestion, that you rewrite the equation as $x^2+ 2x+ 1- 1+ 18= y^2$ so that $(x+ 1)^2= y^2- 17$ is excellent! If you let z= x+1, then that equation becomes $z^2= y^2- 17$ or $y^2- z^2= (y- z)(y+ z)= 17$ . Since y and z are integers, you should be able to quickly find Plato's solution as well as the three other ordered pairs satifying that equation.

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