document.write( "Question 310761: 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...i dont get this \n" ); document.write( "
Algebra.Com's Answer #222306 by scott8148(6628)\"\" \"About 
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let x , x+2 , and x+4 be the integers\r
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\n" ); document.write( "\n" ); document.write( "(x+4)^2 + 65 = x^2 + (x+2)^2\r
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\n" ); document.write( "\n" ); document.write( "x^2 + 8x + 81 = 2x^2 + 4x + 4\r
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\n" ); document.write( "\n" ); document.write( "0 = x^2 - 4x - 77\r
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\n" ); document.write( "\n" ); document.write( "0 = (x-11)(x+7)\r
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\n" ); document.write( "\n" ); document.write( "x = -7 (the given solution)\r
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\n" ); document.write( "\n" ); document.write( "x = 11\r
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\n" ); document.write( "\n" ); document.write( "so the numbers are 11 , 13 , 15
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