Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Find the answer

2. Let $A$ be the set of values of $m$ for which the roots of the equation in $x$
$2(m+1) x^{2}-\left(m^{2}+m+16\right) x+8 m=0$

are both integers. Then $|A|=$
$\qquad$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

2. 2 .

If $m=-1$ , then $x=-\frac{1}{2}$ , which does not meet the requirement. If $m \neq-1$ , then we can get $x_{1}=\frac{m}{2}, x_{2}=\frac{8}{m+1}$ . According to the problem, $m$ is an even number, and $(m+1) \mid 8$ , so $m=0$ or -2.
Therefore, $|A|=2$ .

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