document.write( "Question 621382: Please help me to solve this question:
\n" ); document.write( "Prove that the product of two consecutive odd integers is always 1 less than a multiple of 4. \n" ); document.write( "

Algebra.Com's Answer #390810 by KMST(5328) $\"\"$ $\"About$

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Usually, for problems with consecutive odd numbers, it is enough to call them n and n+2, and the same goes with consecutive even numbers.
\n" ); document.write( "In this case, we have to express the numbers in a way that shows they are odd.
\n" ); document.write( "To specify that a number is even, we can call it $\"2m\"$ and specify that m is a positive integer, so the even number could be 2, 4, 6, etc.
\n" ); document.write( "The numbers right before and right after an even number are consecutive odd integers, so we will use $\"2m-1\"$ and $\"2m%2B1\"$ as our consecutive odd integers.
\n" ); document.write( "Any pair of consecutive odd integers can be expressed as $\"2m-1\"$ and $\"2m%2B1\"$ .
\n" ); document.write( "For any pair of consecutive odd integers we can find their m by calculating their average (the even number between them) and dividing by 2.
\n" ); document.write( "The product of $\"2m-1\"$ and $\"2m%2B1\"$ is
\n" ); document.write( " $\"%282m%2B1%29%282m-1%29=%282m%29%5E2-1%5E2=4m%5E2-1\"$
\n" ); document.write( "Since $\"m\"$ was a positive integer, $\"m%5E2\"$ is a positive integer and
\n" ); document.write( " $\"4m%5E2=4%2Am%5E2\"$ is a multiple of 4.
\n" ); document.write( " $\"4m%5E2-1\"$ is one less than $\"4m%5E2\"$ , so it is 1 less than a multiple of 4. \n" ); document.write( "

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