Cube - Cube = Cube?

Let $x=n+\alpha,$ where $\lfloor x \rfloor=n.$ Then, $0\le \alpha < 1.$

Knowing that $x-\lfloor x \rfloor = n+\alpha-n=\alpha,$ we see that

$(n+\alpha)^3-\lfloor (n+\alpha)^3 \rfloor = \alpha^3 \\\\ n^3+3n^2\alpha+3n\alpha^2+\alpha^3-\lfloor n^3+3n^2\alpha+3n\alpha^2+\alpha^3 \rfloor = \alpha^3 \\\\ n^3+3n^2\alpha+3n\alpha^2 = \lfloor n^3+3n^2\alpha+3n\alpha^2+\alpha^3 \rfloor.$

Since $0\le \alpha^3<1$ and the RHS is an integer, we deduce that $3n^2\alpha+3n\alpha^2$ is an integer. ( $n^3$ is an integer)

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Let $3n^2\alpha+3n\alpha^2=m,$ and solve for $\alpha.$

$\alpha = \dfrac{-3n^2+\sqrt{9n^4+12mn}}{6n} = - \dfrac{n}{2}+\sqrt{\dfrac{n^2}{4}+\dfrac{m}{3n}}.~(\because~\alpha>0)$

Solving $0\le \alpha < 1,$ we find $0\le m < 3n(n+1).$

Therefore there are $3n(n+1)$ possible values for $m.$

Note that if $m$ changes, $x$ changes its value as well.

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The number of $x$ when $1\le n \le k$ is $\displaystyle \sum_{i=1}^{k}3i(i+1)=\dfrac{k(k+1)(2k+1)}{2}+\dfrac{3k(k+1)}{2}=k(k+1)(k+2)<2017.$

Since $11\times12\times13=1716,$ we find that the minimum of $k$ is the $2017-1716=301^{\text{th}}$ number with an integer part of $12.$

The first number with an integer part of $12$ is of course, $12,$ when $m=0.$

This shows that the minimum of $k$ occurs when $m=300.$

$\alpha = - \dfrac{12}{2}+\sqrt{\dfrac{12^2}{4}+\dfrac{300}{3\cdot12}} = -6+\dfrac{\sqrt{399}}{3}.$

Therefore, the minimum of $k$ is $n+\alpha = 12 + \left(-6+\dfrac{\sqrt{399}}{3}\right)=6+\dfrac{\sqrt{399}}{3}.$

$\therefore~a+b+c+d=6+1+399+3=\boxed{409}.$

• Take $1 ≤ x^3 < 2$

We get $x^3 - 1 = (x-1)^3$

This yields exactly one real solution at x = 1

Similar would take place till $6 ≤ x^3 < 7$ and we would get some irrational solutions, but no solution when $7 ≤ x^3 < 8$

Instead one is noticed at $x^3 = 8$ , in the equation : $x^3 - 8 = (x-2)^3$ which has a solution at x = 2

• Observation : In an interval of $a^3 ≤ x < (a+1)^3$ , we have exactly $(a+1)^3 - 1$ solutions ... as illustrated below :

We may also say that the solution at the intervals $7 ≤ x^3 < 8$ and $8 ≤ x^3 < 9$ coincides at x = 8.

As $12^3 = 1728$ and $13^3 = 2197$ , so there will be overall 11 $( x = 2,3,4,.. 12 )$ coinciding cases till 2017

So the $2017^{th}$ solution will exist at $2017 + 11 = 2028$

Thus our equation will become $x^3 - 2028 = (x-12)^3$ which solves to $3x^2 - 36x - 25 = 0$

$\displaystyle x = 6 + 1 \frac{\sqrt{399}}{3}$

• and our answer is 6 + 1 + 399 + 3 = $\boxed {409}$ ​
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