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Find the total number of integral values of a such that $x^2+ax+a+1=0$ has integral roots.

Solution : Finding the discriminant of the question D = $b^2-4ac = a^2-4(a+1)$

= $(a-2)^2-8 $ Let this be equal to $t^2 : \Rightarrow (a-2)^2-8 =t^2 \Rightarrow (a-2)^2-t^2=8 \Rightarrow (a-2-t)(a-2+t)= 8 $

Now how to proceed further in this please guide...

Sachin
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  • Consider two roots $p$ and $q$, and what can you can about the product of roots ? Their sum? And you can even find the difference in terms of $a$. – Inceptio Apr 27 '13 at 13:48

2 Answers2

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Hints:

Suppose $\,\alpha\,,\,\beta\,$ are integer roots of the equation, then by Viete's formulae:

$$\begin{align*}\alpha\beta=&a+1\\\alpha+\beta=&-a\end{align*}$$

This, together with the divisors of $\,8\,$ shall suffice for you to solve.

DonAntonio
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Let us start exactly as you did.

Then note that the last equation you reached says that the product $(a-2-t)(a-2+t)$ is equal to the integer $8$. There are not many possibilities, because $8$ has not many factors. Don't forget about negative factors.

For example, maybe $a-2-t=2$ and $a-2+t=4$. What does that give for $a$? Does it work?

Remark: If instead of $8$ we had, say, $8000000$, we would want to find a quick way to count possibilities without listing.

André Nicolas
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