Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 4"

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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> ?
 
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==
 
==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==
 
==See Also==

Latest revision as of 16:07, 18 December 2018

Problem

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

Solution

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

See Also

2006 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions
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