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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

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  • $\begingroup$ So you have the equation $x^3 + x -a=0, x\in \mathbb{Z}$ $\endgroup$ – user261263 May 24 '16 at 13:39
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$$\lfloor x \rfloor ^3+x-a=0$$ $a\in \mathbb Z \Rightarrow x \in Z$ $$x^3+x=a$$ $$x(x^2+1)=a$$ $a\in[1;1500]$

Then $2\le x\le11$

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

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

...

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

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As you noted, we must have $x $ integer, hence the equation is $$n^{3}+n=a,\, n\in\mathbb{N} $$ and so the problem can be seen as to find $$\sum_{n=1}^{11}\left(n^{3}+n\right) $$ since $11^{3}+11=1342,\,12^{3}+12=1740. $ So $$\sum_{n=1}^{11}\left(n^{3}+n\right)=\sum_{n=1}^{11}n^{3}+\sum_{n=1}^{11}n=4422.$$

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