document.write( "Question 1053784: Find three consecutive positive odd integers such that twice the square of the first one is one less then the product of the second and third \n" ); document.write( "
Algebra.Com's Answer #669180 by jorel555(1290)\"\" \"About 
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Let n be the first odd integer. Then the next integers will be n+2 and n+4. So:
\n" ); document.write( "2nē+1=(n+2)(n+4)
\n" ); document.write( "2nē+1=nē+6n+8
\n" ); document.write( "nē-6n-7=0
\n" ); document.write( "(n-7)(n+1)=0
\n" ); document.write( "n=7, -1
\n" ); document.write( "Your first integer is 7; the next two are 9 and 11. ☺☺☺☺ \n" ); document.write( "
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