#1**+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.

Melody
Nov 2, 2018

#1**+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