Algebraic rooting #1

Quadratic formula: $x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

The discriminant of this equation is $b^2-4ac=a^2-4(a+1)=a^2-4a-4=a^2-4a+4-8=(a-2)^2-8$

For the roots to be an integer, $(a-2)^2-8$ must be a square number.

The only way for this is for the discriminant to equal $1$ .

$\therefore (a-2)^2-8=1$

$(a-2)^2=9$

$a=5$ or $-1$

Let m and n be the roots. Using vieta's formulas, mn = a+1 m+n = -a Sum the equations, mn+m+n = 1 (m+1)(n+1) = 2 Since m and n can be interchanged, m+1 = 1 , n+1 = 2 or m+1 = -1, n+1=-2 From here we can see that there are two values of m+n and thus two values of (-a and thus) a

Sorry for lack of LaTeX style equations