document.write( "Question 344449: Three consecutive odd integers are such that the square of the third integer is 15 greater than the sum of the squares of the first two. One solution is 1, 3, and 5. Find three other consecutive odd integers that also satisfy the given conditions. What are the integers? \n" ); document.write( "

Algebra.Com's Answer #246402 by Fombitz(32388) $\"\"$ $\"About$

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Let N,N+2, and N+4 be the integers.
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\n" ); document.write( " $\"%28N%2B4%29%5E2=N%5E2%2B%28N%2B2%29%5E2%2B15\"$
\n" ); document.write( " $\"N%5E2%2B8N%2B16=N%5E2%2BN%5E2%2B4N%2B4%2B15\"$
\n" ); document.write( " $\"N%5E2%2B8N%2B16=2N%5E2%2B4N%2B19\"$
\n" ); document.write( " $\"N%5E2-4N%2B3=0\"$
\n" ); document.write( " $\"%28N-3%29%28N-1%29=0\"$
\n" ); document.write( "Two solutions:
\n" ); document.write( " $\"N-3=0\"$
\n" ); document.write( " $\"N=3\"$
\n" ); document.write( "The other solution is already known, $\"N=1\"$ .
\n" ); document.write( "The three other integers are $\"3\"$ , $\"5\"$ , and $\"7\"$ .
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\n" ); document.write( "\n" ); document.write( "Verifying,
\n" ); document.write( " $\"7%5E2=3%5E2%2B5%5E2%2B15\"$
\n" ); document.write( " $\"49=9%2B25%2B15\"$
\n" ); document.write( " $\"49=49\"$
\n" ); document.write( "True, good solution.
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