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The positive difference between two consecutive odd perfect squares is 336. Compute the larger of the two squares.

Guest Dec 26, 2018

edited by
Guest
Dec 26, 2018

edited by Guest Dec 26, 2018

edited by Guest Dec 26, 2018

#1**+1 **

We have \(x\) as the smaller odd number and \(x+2\) as the larger odd number.

Translating this into an equation, we have \((x+2)^2-x^2=336\).

Expanding \((x+2)^2-x^2=4x+4\).

Rewriting, we attain \(4x+4=336, 4x=332\) and \(x=83.\)

So, \(x+2=85\) or \(\boxed{7225}.\)

.tertre Dec 26, 2018

#2**0 **

I am sorry for the inconvenience, but your answer, 7225, is incorrect. If you can, please redo this problem or ask for help. This question is way too hard for me. AoPS is a pain in the ____. I want to know if you guys get many questions from that website.

Guest Dec 26, 2018