QUESTION TO ASK ABOUT ?

One root is α and the other is (α + 1).

q
= α(α + 1)
= α² + α

p
= (α + α + 1)
= 2α + 1

p²
= (2α + 1)²
= 4α² + 4α + 1

4q
= 4(α² + α)
= 4α² + 4α

p² - 4q - 1
= 4α² + 4α + 1 -4α² - 4α - 1
= 0

Proven.

Just work out the values of p² and 4q separately, put it in and you're home and dry.

Start from the quadratic equation.

x = ((--p) ± √((-p)^2 - 4*1*q))/(2*1)

x = (p ± √(p^2 - 4q))/2

Lets call the two roots x1 and x2, where x2 is the greater....

x1 = (p - √(p^2 - 4q))/2

x2 = (p + √(p^2 - 4q))/2

x2 - x1 = (p - √(p^2 - 4q))/2 - (p + √(p^2 - 4q))/2 = 1

(p - √(p^2 - 4q) - p - √(p^2 - 4q))/2 = 1

(-2√(p^2 - 4q))/2 = 1

-√(p^2 - 4q) = 1

(p^2 - 4q) = 1

p^2 - 4q - 1 = 0

This so easy , here is the solution:

1.complete the square, this means you will have the following eq.:
x^2-px+q=x^2-px+(p/2)^2-(p/2)^2+q=0.
from which you will have:
(x-p/2)^2=(p/2)^2-q.

2.The solution of this equation will be:
x1=p/2-sqrt((p/2)^2-q), this should be the first number.
x2=p/2+sqrt((p/2)^2-q).

if x2=x1+1 (two consecutive integers) you will get a simple equation , do some manipulations and you will end up with the proof.

you then need [x-n][x-(n+1)] = 0...multiply out , find p & q and show that p² - 4q - 1 = 0...it is not hard