+0  
 
+1
56
1
avatar

Suppose the roots of the polynomial \(x^2 - mx + n\) are positive prime integers (not necessarily distinct). Given that \(m<20\) how many possible values \(n\) of are there?

 Jan 10, 2019

Best Answer 

 #1
avatar+4457 
+2

\(\text{the roots are }\\ x = \dfrac{m \pm \sqrt{m^2-4n}}{2}\)

 

\(\text{as these are integers }m^2 - 4n \text{ must be a perfect square}\\ m<20 \Rightarrow m^2 \leq 19^2 = 361\\ 4n

 

It appears there are indeed 90 valid values of n

 Jan 10, 2019
 #1
avatar+4457 
+2
Best Answer

\(\text{the roots are }\\ x = \dfrac{m \pm \sqrt{m^2-4n}}{2}\)

 

\(\text{as these are integers }m^2 - 4n \text{ must be a perfect square}\\ m<20 \Rightarrow m^2 \leq 19^2 = 361\\ 4n

 

It appears there are indeed 90 valid values of n

Rom Jan 10, 2019

10 Online Users

avatar
avatar