2010/013) Form a quadratic equation whose coefficients are integers and in AP and it has got rational roots

Let the quadratic equation be

ax^2+bx+c = 0


as the coefficients are in ap so let common difference be y so b-a = y c-b = y

or a = b-y and c = b+ y

for the equation to have rational roots we have

discriminant is a perfect square that is say n^2


b^2 – 4 ac = n^2

or b^2-4(b-y)(b+y) = n^2

or b^2-4(b^2 – y^2) = n^2

or 4y^2-3b^2= n^2

dividing by n^2 and letting b/n = t and y/n = s we ge

4s^2-3 t^2= 1 ..1

s and t are real numbers

And from inspection we have a root s = 1/2, t =0 is a rational root and to find another rational root we draw a straight line from (0,1/2)

We get s = mt+ 1/2 is a general root …(2)

So we put in 1 to get the intersection of the curve given in 1 with this straight line to get

(2mt+1)^2 - 3t^2 = 1

Or 4m^2t^2 +4mt+1 – 3t^2 = 1

Or (4m^2-3)t^2 + 4mt = 0

ignoring the starting solution t = 0 we get

t= - 4m/(4m^2-3)

so s= mt+1/2 we get s = -(4m^2-3)/(8m^2-6)

now putting n = 8m^2-6 to get rid of denominator we get

b = -8m and y = -4m^2+3

so we get

a = 4m^2 – 8m + 3
b = - 8m
c = -4m^2 -8m -3 as roots

and taking m as different integers we get different solutions in integers

for example
m = 1 gives (-1.-8, -15) multiply by one we get x^2+ 8x + 15 = 0 giving factor (x+3)(x+5) rational
m = 2 gives (3.-16, -35) we get 3x^2 -16x -35= 0 giving (3x+5)(x-7)
m =3 gives (15,-24,-63) divide by 3 to get 5x^2-8x-21 = (5x+7)x-3)

so on