Make it clearer, please. Is it required to find the cardinality of the set $A\subset \mathbb{R}$, such that for each $a\in A$ the equation $x^3 = ax+a+1$ has a double root? a root that is an even integer? And what does $|x| < 1000$ have to do with all this? EDIT. That correction should make it; now it seems reasonably comprehensible.

mavropnevma wrote: Make it clearer, please. Is it required to find the cardinality of the set $A\subset \mathbb{R}$, such that for each $a\in A$ the equation $x^3 = ax+a+1$ has a double root? a root that is an even integer? And what does $|x| < 1000$ have to do with all this? Statement has been revised to make it more comprehensible.

Wait what? "Determine the number of real $a$ such that for every $a$..." The first part implies that we should fix $a$ but the second has a quantifier.

It's allright; it means to determine the cardinality of the set $A$ of real numbers, such that for each $a\in A$ we have those two conditions fulfilled.

Suppose there exist $x_1$ and $x_2$, such that $\dfrac{x^3_1-1}{x_1+1}=\dfrac{x^3_2-1}{x_2+1}$, where $x_1,x_2$ are distinct even numbers. then $(x_1-x_2)(x^2x_2+x_1x^2_2+x^2_1+x^2_2+x_1x_2+1)=0$ . Since $x_1\not= x_2$, hence $x_1-x_2\not= 0$. Besides, $x^2x_2+x_1x^2_2+x^2_1+x^2_2+x_1x_2$ is even, thus $(x_1-x_2)(x^2x_2+x_1x^2_2+x^2_1+x^2_2+x_1x_2+1)\not=0$, which is a contradiction. So for arbitrary $|x|\le 998$, x is even, $a=\dfrac{x^3-1}{x+1}$ obtains distinct value. Hence there are $999$ such real number $a$ satisfying the conditions.

jred wrote: Besides, $x^2x_2+x_1x^2_2+x^2_1+x^2_2+x_1x_2$ is even, thus $(x_1-x_2)(x^2x_2+x_1x^2_2+x^2_1+x^2_2+x_1x_2+1)\not=0$, which is a contradiction. Note that $x^2x_2+x_1x^2_2+x^2_1+x^2_2+x_1x_2+1=(x_1+1)(x_2+1)(x_1+x_2-1)+2$ is actually nonzero for any integers $x_1,x_2$ (otherwise there'll be exactly $2$ odd numbers among $x_1+1,x_2+1,x_1+x_2-1$, which is impossible).