View source for 2006 UNCO Math Contest II Problems/Problem 4

== Problem ==

Determine all positive integers <math>n</math> such that <math>n^2+3</math> divides evenly (without remainder) into <math>n^4-3n^2+10</math> ?

==Solution==
By polynomial division, <math>\frac{n^4 - 3n^2 + 10}{n^2 + 3} = n^2 - 6 + \frac{28}{n^2 + 3}</math>. By default, <math>n^2 - 6 \in \mathbb{N}</math> when <math>n \in \mathbb{N}</math>, so we must find all positive integral values of <math>n</math> for which <math>\frac{28}{n^2 + 3}</math> is also an integer. Listing out the factors of <math>28</math>, we see that <math>n^2 + 3 \in \{1, 2, 4, 7, 14, 28\}</math>. Looking for positive integral values that satisfy these conditions yields <math>\boxed{n \in \{1, 2, 5\}}</math>, as desired. <math>\blacksquare</math>

==See Also==
{{UNCO Math Contest box|n=II|year=2006|num-b=3|num-a=5}}

[[Category:Introductory Algebra Problems]]