SOLUTION: 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?

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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?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the smallest odd integer.
Since odd integers are two apart from each other (think about it), the next two consecutive odd integers would be:
x+2
x+2+2=x+4

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

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+=+3%28x%5E2+%2B+4x+%2B+7%29
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.

So the integer factors of n are:
  • 3 and x%5E2%2B4x%2B7
  • 1 and n, of course

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.