Question 227525
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<br>
So
{{{n = 1 + (x)^2 + (x+2)^2 + (x+4)^2}}}
Simplifying this, using FOIL or the pattern {{{(a+b)^2 = a^2 +2ab + b^2}}} on the binomials:
{{{n = 1 + (x^2) + (x^2 + 4x + 4) + (x^2 + 8x + 16)}}}
Adding like terms:
{{{n = 3x^2 +12x + 21}}}<br>
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(x^2 + 4x + 7)}}}
Next we try other factoring techniques on {{{x^2 + 4x +7}}}: patterns, trinomial factoring, factoring by grouping, and trial and error of the possible rational roots. However {{{x^2+4x+7}}} will not factor with any of these methods.<br>
So the integer factors of n are:<ul><li>3 and {{{x^2+4x+7}}}</li><li>1 and n, of course</li></ul>
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 <b>all</b> such n's is 3.