document.write( "Question 897396: Three consecutive odd integers are such that the square of the third integer is 65 less than the sum of the squares of the first two. One solution is -7, -5, and -3. Find three other consecutive odd integers that also satisfy the given conditions. \n" ); document.write( "

Algebra.Com's Answer #544697 by Fombitz(32388) $\"\"$ $\"About$

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Let $\"N-2\"$ , $\"N\"$ , and $\"N%2B2\"$ be the integers.
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\n" ); document.write( "\n" ); document.write( " $\"%28N%2B2%29%5E2=N%5E2%2B%28N-2%29%5E2-65\"$
\n" ); document.write( " $\"N%5E2%2B4N%2B4=N%5E2%2BN%5E2-4N%2B4-65\"$
\n" ); document.write( " $\"N%5E2-8N-65=0\"$
\n" ); document.write( " $\"%28N-13%29%28N%2B5%29=0\"$
\n" ); document.write( "Two solutions:
\n" ); document.write( " $\"N-13=0\"$
\n" ); document.write( " $\"N=13\"$
\n" ); document.write( "The integers are 11,13,and 15.
\n" ); document.write( "The other solution is already given.\r
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