Given $a$, $b$ are integers with $a > b$ and the two roots $\alpha$, $\beta$ of the equation $3x^2 + 3(a+b)x + 4ab = 0$ satisfy a relation

Question

Given $a$, $b$ are integers with $a > b$ and the two roots $\alpha$, $\beta$ of the equation $3x^2 + 3(a+b)x + 4ab = 0$ satisfy the relation $$\alpha(\alpha + 1) + \beta(\beta + 1) = (\alpha + 1)(\beta + 1)$$, find all the pairs $(a, b)$ of two integers.

Greetings, I was doing the above question related to number theory. I could not complete the question.

Here's what I've done so far:

Note that $\alpha + \beta = -(a + b)$ and $\alpha \beta = \frac{4ab}{3}$. Now expanding $\alpha(\alpha + 1) + \beta(\beta + 1) = (\alpha + 1)(\beta + 1)$ gives us $\alpha^2 + \beta^2 - \alpha \beta = 1$. Thus, $(\alpha + \beta)^2 - 3\alpha \beta = 1$. Substituting and doing some calculations, we get $a - b = 1$.

Now I can't proceed further. I don't know if this problem has been posted earlier on MSE (I have checked though).

Any help would be appreciated.

Thank You

Answer 1

Now, we have $b=a-1$ and $$3x^2+3(2a-1)x+4a(a-1)=0,$$ which gives $$9(2a-1)^2-4\cdot3\cdot4a(a-1)\geq0.$$ Can you end it now?

Answer 2

Using $b=a-1$, the equation is:

$$3x^2+3(2a-1)x+4a(a-1)=0$$

The discriminant

$$9+12a-12a^2$$

is positive only for the integers $a=0$ and $a=1$