## Saturday, August 15, 2015

### Find the sum of all possible $a^3$, where $a$ is a rational figure (Heuristic Solution)

Given that $a$ is rational and the equation $ax^2+(a+2)x+a-1=0$ has integer roots.

Find the sum of all possible $a^3$.

It is very important for me to say out loud here that this solution is provided by my math friend, a retired math professor from the U.K.

Picking up where we left, where it is useless if we keep the $a$, a rational number in the given quadratic equation.

But it can't be that we could just remove $a$ and make it disappear into of thin air, we have to devise a good plan for it.

Note that we can divide through the quadratic equation by $a$, and get:

$ax^2+(a+2)x+a-1=0$

$x^2+\left(1+\dfrac{2}{a}\right)x+1-\dfrac{1}{a}=0$

Now we can see it clearly that

the sum of the two integer roots $=-\left(1+\dfrac{2}{a}\right)$

the product of the two integer roots $=1-\dfrac{1}{a}$

Note that both must yield another integer figure, therefore, we let  $\dfrac{1}{a}=b$, where $b$ is an integer.

Now, we replace $\dfrac{1}{a}$ by $b$ into $x^2+\left(1+\dfrac{2}{a}\right)x+1-\dfrac{1}{a}=0$, we have:

$x^2+\left(1+2b\right)x+1-b=0$

Therefore

\begin{align*}\text{Discriminant}&=(1+2b)^2-4(1)(1-b)\\&=1+4b+4b^2-4+4b\\&=4b^2+8b-3\\&=4(b^2+2b)-3\\&=4(b+1)^2-3-4\\&=4(b+1)^2-7\end{align*}

If $a=0$ then the equation becomes $2x-1=0$, which does not have a n integer solution. So $a\ne0$ and we can divide through by $a$, getting

$x^2+\left(1+\dfrac{2}{a}\right)x+1-\dfrac{1}{a}=0$

The product of the roots is $1 - \dfrac{1}{a}$, which must be an integer, so $\dfrac{1}{a}$ is an integer, say $a= \dfrac{1}{n}$.

The equation is then $x^2 + (1+2n)x + 1-n = 0$ and its roots are $x = \frac12\bigl(-1-2n \pm\sqrt{(1+2n)^2 - 4(1-n)}\bigr).$

The discriminant is $(1+2n)^2 - 4(1-n)= 4n^2 + 8n - 3 = 4(n+1)^2 - 7,$ and this must be a square, say $4(n+1)^2 - 7 = m^2.$

But the only squares that differ by $7$ are $9$ and $16$. It follows that $4(n+1)^2 = 16$, so $n+1 = \pm 2.$ Therefore $n=1$ or $-3$, and $a=1$ or $-\frac13.$ The sum of cubes is therefore $1 - \frac1{27} = \frac{26}{27}.$