We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
243
1
avatar+159 

given the equation x^2 - 2px + p^2 -5p - 1 = 0 which has 2 integer solutions

if p is a prime number, what are the values for p?

 

edit: thank you so much!

 Nov 2, 2018
edited by hearts123  Nov 2, 2018

Best Answer 

 #1
avatar+103674 
+3

I just found two, p=3 and  p=7 both work. There may well be many more.

 

\(x^2 - 2px + (p^2 -5p - 1) = 0\\ x=\frac{2p\pm\sqrt{4p^2-4(p^2-5p-1)}}{2}\\ x=\frac{2p\pm\sqrt{20p+4}}{2}\\ x=\frac{2p\pm2\sqrt{5p+1}}{2}\\ x=p\pm\sqrt{5p+1}\\ \text{since x must have 2 integer solutions, 5p+1 must be a perfect square.} \)

 

 

If p=3  then 5p+1=16 which is a perfect square.

If p=7 then  5p+1=36 which is a perfect square.

 Nov 2, 2018
 #1
avatar+103674 
+3
Best Answer

I just found two, p=3 and  p=7 both work. There may well be many more.

 

\(x^2 - 2px + (p^2 -5p - 1) = 0\\ x=\frac{2p\pm\sqrt{4p^2-4(p^2-5p-1)}}{2}\\ x=\frac{2p\pm\sqrt{20p+4}}{2}\\ x=\frac{2p\pm2\sqrt{5p+1}}{2}\\ x=p\pm\sqrt{5p+1}\\ \text{since x must have 2 integer solutions, 5p+1 must be a perfect square.} \)

 

 

If p=3  then 5p+1=16 which is a perfect square.

If p=7 then  5p+1=36 which is a perfect square.

Melody Nov 2, 2018

20 Online Users

avatar
avatar
avatar
avatar