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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+3583 
+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+3583 
+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

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