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

 Dec 26, 2018
edited by Guest  Dec 26, 2018
edited by Guest  Dec 26, 2018
 #1
avatar+4221 
+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 \(\boxed{7225}.\)

.
 Dec 26, 2018
 #2
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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. 

 Dec 26, 2018
 #3
avatar+4221 
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??? I'll check my work again, are they asking for the 85 or 7225...

tertre  Dec 26, 2018
 #4
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The larger of the two squares

 Dec 26, 2018
 #5
avatar+4221 
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85^2-83^2=336, so it is true..

 

Or, (a+b)(a-b)=(85+83)(85-83)=168*2=336.

 Dec 26, 2018
edited by tertre  Dec 26, 2018
 #6
avatar+100513 
+1

Notice that tertre's answer of  83 and 85  is correct

 

85^2 - 83^2  =

 

7225  - 6889   =

 

336.....

 

Good job, tertre!!!!

 

 

cool cool cool

 Dec 26, 2018
edited by CPhill  Dec 26, 2018
edited by CPhill  Dec 26, 2018
 #7
avatar+4221 
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Thank you, CPhill! smiley

tertre  Dec 26, 2018
 #8
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Ok thanks.

 Dec 26, 2018
 #9
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Do you guys deal with a bunch of AoPS problems like this?

 Dec 26, 2018
 #10
avatar+4221 
+1

I do AoPs, and we get a lot of problems, but it helps!

tertre  Dec 26, 2018

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