SOLUTION: Three consecutive odd integers are such that the square of the third is 264 more than the square of the second. Find the three integers.

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Question 169539: Three consecutive odd integers are such that the square of the third is 264 more than the square of the second. Find the three integers.
Found 2 solutions by Mathtut, Edwin McCravy:
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
call the integers a,a+2,a+4
%28a%2B4%29%5E2=%28a%2B2%29%5E2%2B264
a%5E2%2B8a%2B16=a%5E2%2B4a%2B4%2B264
4a=252
highlight%28a=63%29 so the integers are 63,65,67

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Three consecutive odd integers are such that the square of the third is 264 more than the square of the second. Find the three integers. 

First odd integer  = x
Second odd integer = x%2B2
Third odd integer  = x%2B4

the square of the third = %28x%2B4%29%5E2

the square of the second = %28x%2B2%29%5E2

>>...the square of the third is 264 more than the square of the second...<<

Replace "the square of the third" by %28x%2B4%29%5E2
 
>>...%28x%2B4%29%5E2 is 264 more than the square of the second...<<

Replace "the square of the second" by %28x%2B2%29%5E2.

>>...%28x%2B4%29%5E2 is 264 more than %28x%2B2%29%5E2...<<

Replace the word "is" by %22=%22:

>>...%28x%2B4%29%5E2 %22=%22 264 more than %28x%2B2%29%5E2...<<

Make %28x%2B2%29%5E2 264 more than it is by adding 264 to it on the right:

That is, replace "264 more than %28x%2B2%29%5E2" by %28x%2B2%29%5E2%2B264

>>...%28x%2B4%29%5E2 %22=%22 %28x%2B2%29%5E2%2B264...<<

So the equation is

%28x%2B4%29%5E2=%28x%2B2%29%5E2%2B264

Can you solve that equation?  If not post again asking how.

Solution to that equation: x=63

So:

First integer  = x=63
Second integer = x%2B2=63%2B2=65
Third integer  = x%2B4=63%2B4=67

Answer:  63, 65, 67

Edwin