Find the number of positive real values of $a$, such that $a^2 x^2 + ax - 1 + 7a^2$ has $2$ distinct integer roots

Question

Problem Find the number of positive real values of $a$ for which the given equation has 2 distinct integer roots.

$$ a^2 x^2 + ax - 1 + 7a^2 $$

Till now, I have the following inferences. Assuming $\alpha, \beta$ to be the roots, we have

$$\alpha + \beta = \frac{-1}{a}$$ $$\alpha\beta = -\frac{1}{a^2} + 7$$

But since $\alpha, \beta$ are integers, this implies that $|a| \leq 1$. Also by the quadratic formula, we have

$\alpha = \frac{-1 \pm \sqrt{28a^2 - 3}}{2a}$

From here we can infer that since the requirement is just integer roots, if $a$ satisfies the condition, then $-a$ also satisfies the condition. Thus, we may find the total number of solutions and divide by two to get the number of positive integers. However, whether this is helpful or not, I can not figure. One trivial solution emerges, i.e. $a = \pm 1$. I cannot make progress beyond this.

Answer 1

$$a^2x^2+ax+(7a^2-1)=0$$

Let $m,n$ be integer roots of the quadratic . Then, you have :

$$ \begin{align}m+n&=-\frac 1a, \thinspace a≠0\\ mn&=7-\left(\frac {1}{a}\right)^2\end{align} $$

This leads to :

$$ \begin{align}(m+n)^2+mn&=7\\ m^2+3mn+(n^2-7)&=0\end{align} $$

Thus, we can assume that :

$$\Delta_m=5m^2+28=k^2, \thinspace k\in\mathbb Z$$

Working with $\color{#c00}{\rm{mod}\;5}$, we have $k=5p+q$ , where $1≤q≤4$, which yields $5\mid q^2-3$ . A contradiction .

Therefore, no solution exists .

Answer 2

I guess the question is wrong according to your answer that is 3 , this question has been taken from 16-th All-Russian Mathematical Olympiad 1990 round 4 grade 9 P5 .

The original question is Find all positive values of a for which both roots of the equation ($a^2$)($x^2$) + ax + 1 -7($a^2$)=0 are integer . So I will answer according to the original question .

Rearrange the equation by taking quadratic in a (Treat x as constant instead of a)

so ($a^2$)[$x^2$ - 7] + ax + 1 =0

as both a and x are integer therefore D = k^2 (for any integer k)

so $x^2$ -4($x^2$ - 7)*1 =$k^2$

= 28 - 3$x^2$ = $k^2$

clearly only x = -1,-2,-3,1,2,3 satisfy the condition

Now we have to check by placing the possible values of x in quadratic formula if a is also an integer or not .

checking it we will find it that (a,b) = (-1,2) ,(-1,3),(1,-2) satisfy .