Calculate $ \left \lfloor \frac{2017^{3}}{2015 \cdot 2016} - \frac{2015^{3}}{2016 \cdot 2017} \right \rfloor $

Question

Calculate $$ \left \lfloor \frac{2017^{3}}{2015 \cdot 2016} - \frac{2015^{3}}{2016 \cdot 2017} \right \rfloor $$

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attempt:

$$ \frac{2017^{3}}{2015 \cdot 2016} - \frac{2015^{3}}{2016 \cdot 2017} = \frac{2017^{4} - 2015^{4}}{2015 \cdot 2016 \cdot 2017} $$ $$ \frac{(2017^{2} - 2015^{2})(2017^{2} + 2015^{2})}{2015 \cdot 2016 \cdot 2017} = \frac{2(4032)(2017^{2} + 2015^{2})}{2015 \cdot 2016 \cdot 2017} $$ $$ =\frac{4(2017^{2} + 2015^{2})}{2015 \cdot 2017} = \frac{4 \cdot 2017}{2015} + \frac{4 \cdot 2015}{2017} $$

$$ = \frac{8068}{2015} + \frac{8060}{2017} = \frac{8060 + 8}{2015} + \frac{8068 - 8}{2017} $$ $$ = 8 + \frac{8}{2015} - \frac{8}{2017} $$

So the simplified value is 8. Are there more simpler ways?

Answer 1

$$4\cdot\dfrac{(a+1)^2+(a-1)^2}{a^2-1}$$

$$=8\cdot\dfrac{a^2-1+2}{a^2-1}$$

$$=8+\dfrac{16}{a^2-1}$$

Now $0\le\dfrac{16}{a^2-1}<1$ if $a^2-1> 16\iff a^2>17$

Answer 2

Alternatively: $$\left \lfloor \frac{2017^{3}}{2015 \cdot 2016} - \frac{2015^{3}}{2016 \cdot 2017} \right \rfloor=\\ \left \lfloor \frac{2017^3-1+\color{red}1}{2015 \cdot 2016} - \frac{2015^{3}+1-\color{blue}1}{2016 \cdot 2017} \right \rfloor=\\ \left \lfloor \frac{2017^2+2017+1}{2015} - \frac{2015^{2}-2015+1}{2017}+\frac{\color{red}1}{2015}-\frac{\color{blue}1}{2017} \right \rfloor=\\ \small\left \lfloor \frac{(2015+2)^2+(2015+2)+1}{2015} - \frac{(2017-2)^{2}-(2017-2)+1}{2017}+\frac{\color{red}1}{2015}-\frac{\color{blue}1}{2017} \right \rfloor=\\ \left \lfloor 2015+4+1+\frac{7}{2015}-2017+4+1- \frac{7}{2017}+\frac{\color{red}1}{2015}-\frac{\color{blue}1}{2017} \right \rfloor=\\ \left \lfloor 8+\frac{8}{2015}-\frac{8}{2017} \right \rfloor=\\$$

Answer 3

I would avoid any calculations unless I utterly need them

$\left \lfloor \frac{2017^{3}}{2015 \cdot 2016} - \frac{2015^{3}}{2016 \cdot 2017} \right \rfloor=$

$\left \lfloor \frac{2017^{4}}{2015 \cdot 2016\cdot 2017} - \frac{2015^{4}}{2015\cdot2016 \cdot 2017} \right \rfloor=$

$ \lfloor \frac 1{{(2016-1)\cdot2016 \cdot (2016+1)}}[(2016+1)^4 - (2016-1)^4]\rfloor$

Now $(2016\pm 1)^4 = 2016^4 \pm 4*2016^3 + 6*2016^2 \pm 4*2016 + 1$

So $(2016+1)^{4}-(2016-1)^{4}$ will equal $8*2016^3 + 8*2016$.

And so our calculation continues:

$ \lfloor \frac 8{(2016^2-1)2016}[2016^3 + 2016]\rfloor=$

$ \lfloor \frac 8{2016^2-1}[2016^2 + 1]\rfloor=$

$\lfloor 8*(\frac{2016^2 +1}{2016^2 -1})\rfloor=$

$\lfloor 8(\frac{2016^2-1 +2}{2016^2 -1})\rfloor=$

$\lfloor 8(1+ \frac 2{2016^2 -1})\rfloor=$

$\lfloor 8 + \frac {2*8}{2016^2-1}\rfloor$

Now hopefully we can convince ourselves that $\frac {2*8}{2016^2-1}<1$ without actually having to calculate anything.

So the result is $8$.

Answer 4

Let $a = 2016$, the expression inside the floor is

$$\begin{align} \frac{(a+1)^3}{a(a-1)} - \frac{(a-1)^3}{a(a+1)} = & \frac{(a+1)^4 - (a-1)^4}{a(a^2-1)}\\ = & \frac{(a^4+4a^3+6a^2+4a+1)-(a^4-4a^3+6a^2-4a+1)}{a(a^2-1)}\\ = &\; 8\frac{a^3+a}{a(a^2-1)} = 8\frac{a^2+1}{a^2-1} = 8 + \frac{16}{a^2-1}\end{align} $$ It is clear $0 < \frac{16}{a^2-1} < 1$, this means the answer is $8$.