<s>[INST] Value of $a$ such that the equation has integral roots If $(a+b+c)x^2+2(a-b)x +a+b-8=0$ has integral roots and $a,b,c \in R^+ ,c>b-3a>0$,then the value of $a$ is ----
I am aware that a condition for integral roots is that the discriminant must be a perfect square but I couldn't deduce anything from it.Any ideas?Thanks.
 [/INST] Rewrite as 
$$\tag1 x^2+px+q = 0$$
with $$ p=\frac{2(a-b)}{a+b+c},\qquad q=\frac{a+b-8}{a+b+c}.$$
As $p$ is the negative sum and $q$ is the product of the roots (Vieta), both must be integers.
From the given inequality, 
$$-p=\frac{(a+b)+(b-3a)}{a+b+c}\begin{cases}<\frac{a+b+c}{a+b+c}=1\\
>\frac{a+b}{a+b+c}>0 \end{cases}.$$
But there is no such integer.
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