MHB Quadratic equation with rational roots

[kaliprasad]

form quadratic equation $ax^2 +bx+c=0$ in parametric form such that a,b,c are integers in AP and it has got rational roots

 

[Opalg]

I worked this out when thinking about http://mathhelpboards.com/challenge-questions-puzzles-28/forming-quadratic-equation-10437.html.

[sp]Let $b = a+k$ and $c = a+2k$. Then the equation is $ax^2 + (a+k)x + (a+2k) = 0$, and the solutions are $x = \frac1{2a}\bigl( -a-k \pm\sqrt{(a+k)^2 - 4a(a+2k)}\bigr).$ The condition for rational roots is that the expression under the square root should be a square, namely $(a+k)^2 - 4a(a+2k) = \Box.$ Write that as $(k-3a)^2 - 3(2a)^2 = \Box$. This shows that we are looking for integer solutions of the equation $x^2 + 3y^2 = z^2.$

That is similar to the Pythagorean triples equation. As in that equation, we look for primitive solutions (those in which $x$, $y$ and $z$ are co-prime). The method is similar to the Pythagorean case, and the parametric formula for the solution is that for any co-prime integers $m$, $n$, with $m$ not a multiple of $3$, we have $x=3n^2 - m^2$, $y = 2mn$ and $z = 3n^2+m^2$. In terms of $a$ and $k$, that gives $a = mn$, $k = m^2+3mn+3n^2$. Finally, if we substitute into the expressions for $b$ and $c$, we see that the primitive solutions are $$a = mn,$$ $$b = m^2 + 4mn + 3n^2,$$ $$c = 2m^2 + 7mn + 6n^2,$$ where (as before) $m$ and $n$ are co-prime and $m$ is not a multiple of $3$.[/sp]

 

[Albert1]

form quadratic equation $ax^2 +bx+c=0$ in parametric form such that a,b,c are integers in AP and it has got rational roots

Spoiler

let $a=b-d, c=b+d$
if it has got rational roots
we have :$b^2-4(b+d)(b-d)=b^2-4(b^2-d^2)=4d^2-3b^2=k^2$
here $b,d,k\in Z$
there are many quadratic equatons meet the restriction
in particular if $c=(b+d)=0 \,\,or\,\,(d=-b)$the equation will be:
$2bx^2+bx=0$
and the solutons are $x=0,$ and $x=\dfrac {-1}{2}$
if $b=0 \,\,and\,\,(d\neq 0)$the equation will be:
$-dx^2+d=0$
and the solutons are $x=\pm1$

 

[Opalg]

This is to reconcile Albert's solution with mine.

[sp]I defined a primitive solution to be one in which the coefficients $a$, $b$, $c$ have no common factor (other than $\pm1$). In Albert's family of equations $2bx^2 + bx = 0$, the only primitive solutions are those where $b = \pm1.$ In particular, if $b = -1$ then the equation becomes $(-2)x^2 + (-1)x + 0 = 0$, which occurs in my family of primitive solutions by taking $m= -2$ and $n=1$.

That example raises something that I overlooked in my solution. My formula for the common difference $k$ in the AP is $k = m^2 + 3mn + 3n^2$. That expression is positive definite, so my formula only covers equations whose coefficients form an AP in which the common difference is positive. But if $ax^2 + bx + c = 0$ has coefficients forming an AP with negative common difference then by taking its negative you get the equation $(-a)x^2 + (-b)x + (-c) = 0$. That is obviously just a different way of writing the same equation, but this time the coefficients form an AP with positive common difference.

So I need to reformulate the definition of a primitive solution to be one in which $a$, $b$, $c$ have no common factor (other than $\pm1$) and the common difference $b-a$ is positive. Then the formula $$a = mn,$$ $$b = m^2 + 4mn + 3n^2,$$ $$c = 2m^2 + 7mn + 6n^2,$$ where $m$ and $n$ are nonzero co-prime integers, $m$ is not a multiple of $3$, and $n>0$, gives what the OP wanted, namely a parametric form for the primitive solutions. The general solution is then obtained by multiplying all the coefficients in a primitive solution by some nonzero integer $d$.[/sp]

 

[kaliprasad]

my solution

Spoiler

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 $\frac{b}{n} = t$ and $\frac{y}{n}$ = s we get

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

s and t are real numbers

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

We get $s = mt+ \frac{1}{2}$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

$t= \frac{- 4m}{4m^2-3}$

so $s = mt+ \frac{1}{2}$ we get $s = -\frac{4m^2-8m + 3}{8m^2-6}$

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

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

so we get

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

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