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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.

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  • $\begingroup$ Do you require the $a$ of your starting polynomial to be an integer, or only the roots of the polynomial? $\endgroup$
    – coffeemath
    Oct 6, 2023 at 9:27
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    $\begingroup$ $a$ should be real, I have edited the question. $\endgroup$ Oct 6, 2023 at 9:32
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    $\begingroup$ $a=-1/\gamma$, where $\gamma=\alpha+\beta$ is a positive integer. From there, perhaps you can get to $28-3\gamma^2$ being a perfect square, which should narrow things down. $\endgroup$ Oct 6, 2023 at 9:39
  • $\begingroup$ Check your formula for $\alpha\beta$ again. It should be $7-\frac1{a^2}$ instead of $\frac1{a^2}-7$ for your quadratic. $\endgroup$ Oct 6, 2023 at 10:46
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    $\begingroup$ I’m voting to close this question because OP is not responding. $\endgroup$ Oct 10, 2023 at 11:07

2 Answers 2

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

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 .

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  • $\begingroup$ The answer given for the problem was 3. $\endgroup$ Oct 6, 2023 at 10:43
  • $\begingroup$ @ShivanshJaiswal Well, then let's wait for feedback from mathematicians. Unfortunately, I'm not an expert .. $\endgroup$ Oct 6, 2023 at 10:45
  • $\begingroup$ I think the answer given is wrong, your answer seems correct. $\endgroup$ Oct 7, 2023 at 4:41
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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 .

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