```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.

<pre><font size = 4 color = "indigo"><b>
First odd integer  = {{{x}}}
Second odd integer = {{{x+2}}}
Third odd integer  = {{{x+4}}}

the square of the third = {{{(x+4)^2}}}

the square of the second = {{{(x+2)^2}}}

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

Replace "the square of the third" by {{{(x+4)^2}}}

>>...{{{(x+4)^2}}} is 264 more than the square of the second...<<

Replace "the square of the second" by {{{(x+2)^2}}}.

>>...{{{(x+4)^2}}} is 264 more than {{{(x+2)^2}}}...<<

Replace the word "is" by {{{"="}}}:

>>...{{{(x+4)^2}}} {{{"="}}} 264 more than {{{(x+2)^2}}}...<<

Make {{{(x+2)^2}}} 264 more than it is by adding 264 to it on the right:

That is, replace "264 more than {{{(x+2)^2}}}" by {{{(x+2)^2+264}}}

>>...{{{(x+4)^2}}} {{{"="}}} {{{(x+2)^2+264}}}...<<

So the equation is

{{{(x+4)^2=(x+2)^2+264}}}

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+2=63+2=65}}}
Third integer  = {{{x+4=63+4=67}}}