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If $\alpha ,\beta$ be the integer roots of the quadratic equation $ax^{2}+bx+c=0$ where $a,b,c$ are positive and in AP in that order then find the value of $\alpha + \beta +\alpha \beta$.

My attempt $:$

Before going to find the above value let us first prove the following lemma.

Lemma :

There exists no positive integer $n$ other than $7$ such that $n^2 -6n -3$ is a perfect square.

Proof $:$

First we observe that $n^2-6n-3 < n^2$ and $n^2-6n-3 \ne 0$ if $n$ is a positive integer.So if $n^2-6n-3$ is a perfect square then there exists a positive integer $m$ with $m<n$ such that $n^2-6n-3 = (n-m)^2=n^2-2mn+m^2$ $\implies$ $n= \frac {m^2+3} {2(m-3)}$ $\implies$ $2\not|\ m$ and $m >3$ since $n>0$. So $n-m =\frac {3+6m-m^2}{2(m-3)}$. Since $n-m>0$ then we have $m^2<6m+3$.So $m \leq 6$.Now since $2 \not|\ m$ and $m>3$ we only have one option left which is $m=5$ and hence $n=7$.This proves the lemma.

Now we proceed to find the value of $\alpha + \beta + \alpha \beta$.

Since $a,b,c$ are in AP we may take $b=a+r$ and $c=a+2r$.Then by the relation between roots and coefficients we have

$\alpha + \beta =-\frac {a+r} {a}$ and $\alpha \beta = \frac {a+2r} {a}$.So we have $\alpha + \beta +\alpha \beta = \frac {r} {a}$.This shows that $a \mid r$.So $r=na$ for some integer $n$.But here we should observe a fact that $n>0$ for otherwise all of $a,b,c$ would not be positive.Since the given quadratic equation has integer roots so the discriminant of this quadratic should be a perfect square.Now the discriminant of the above quadratic is $(a+r)^2 - 4a(a+2r)=r^2-6ar-3a^2$.Now putting $r=na$ we have $(n^2-6n-3)a^2$ which is a perfect square if $n^2-6n-3$ is a perfect square.So by the above lemma the above quadratic equation has integer roots if $n=7$.So $r=7a$ and hence $\alpha + \beta + \alpha \beta=7$.

Is the above method correct at all?Please verify it.Also mention easier method in finding the above value (if any).

Thank you in advance.

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