document.write( "Question 227525: A number n is 1 plus the sum of the squares of three consecutive odd integers. What is the largest integer factor of all such numbers n? \n" ); document.write( "

Algebra.Com's Answer #169041 by jsmallt9(3758) $\"\"$ $\"About$

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Let x = the smallest odd integer.
\n" ); document.write( "Since odd integers are two apart from each other (think about it), the next two consecutive odd integers would be:
\n" ); document.write( "x+2
\n" ); document.write( "x+2+2=x+4

\n" ); document.write( "So
\n" ); document.write( " $\"n+=+1+%2B+%28x%29%5E2+%2B+%28x%2B2%29%5E2+%2B+%28x%2B4%29%5E2\"$
\n" ); document.write( "Simplifying this, using FOIL or the pattern $\"%28a%2Bb%29%5E2+=+a%5E2+%2B2ab+%2B+b%5E2\"$ on the binomials:
\n" ); document.write( " $\"n+=+1+%2B+%28x%5E2%29+%2B+%28x%5E2+%2B+4x+%2B+4%29+%2B+%28x%5E2+%2B+8x+%2B+16%29\"$
\n" ); document.write( "Adding like terms:
\n" ); document.write( " $\"n+=+3x%5E2+%2B12x+%2B+21\"$

\n" ); document.write( "Now we can find the factors of n. When factoring, always start with the Greatest Common Factor (GCF). In this case the GCF is 3:
\n" ); document.write( " $\"n+=+3%28x%5E2+%2B+4x+%2B+7%29\"$
\n" ); document.write( "Next we try other factoring techniques on $\"x%5E2+%2B+4x+%2B7\"$ : patterns, trinomial factoring, factoring by grouping, and trial and error of the possible rational roots. However $\"x%5E2%2B4x%2B7\"$ will not factor with any of these methods.

\n" ); document.write( "So the integer factors of n are:

3 and $\"x%5E2%2B4x%2B7\"$
1 and n, of course

\n" ); document.write( "Of these factors only 1 and 3 do not depend, directly or indirectly, on the value of x. And the larger of these is 3. So the largest integer factor of all such n's is 3. \n" ); document.write( "

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