[Math] sum of all positive integral values of $a\;,$ for which equation $\lfloor x \rfloor ^3+x-a=0$ has solution

Question

The sum of all positive integral values of $a\;,$ Where $a\in \left[1,1500\right]$

for which the equation $\lfloor x \rfloor ^3+x-a=0$ has solution, Where $\lfloor x \rfloor $ Represent floor of $x$

$\bf{My\; Try::}$ Given $\lfloor x \rfloor ^3+x-a=0\Rightarrow x=\underbrace{a-\lfloor x \rfloor^3}_{\bf{integer\; quantity}}$

So Here $x$ must be an $\bf{Integer\; quantity.}$

Now How can I solve after that, Help Required, Thanks

Answer 1

$$\lfloor x \rfloor ^3+x-a=0$$ $a\in \mathbb Z \Rightarrow x \in Z$ $$x^3+x=a$$ $$x(x^2+1)=a$$ $a\in[1;1500]$

Then $2\le x\le11$

If $x=2$ then $a=10$

If $x=3$ then $a=3\cdot10=30$

...

If $x=11$ then $a=11\cdot(11^2+1)=1342$