<s>[INST] Sum of integer reciprocals The following problem results in a cubic, but is there a way to simplify it easier?:
The sum of the reciprocals of three consecutive integers is 47/60. What is the sum of these integers? 
I ended up with $47{ x }^{ 3 }-180{ x }^{ 2 }-47x+60 =0$. However, unlike other methods where I've learned to "Let ${ x }^{ 2 }$ = k, $\therefore  { x }^{ 4 }={ k }^{ 2 }$", ${x}^{3}$ is not a multiple of 2, so I cannot apply this same method. How should I proceed? (Other than just substitute, which the answers from this & this website did).
 [/INST] This answer is based upon symmetry and a guess.

We try to exploit symmetry and use the approach
  \begin{align*}
\color{blue}{\frac{47}{60}}&=\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}\\
&=\frac{n(n+1)+n^2-1+(n-1)n}{n(n^2-1)}\\
&\color{blue}{=\frac{3n^2-1}{n(n^2-1)}}
\end{align*}

Let's take a closer look at 
\begin{align*}
\frac{3n^2-1}{n(n^2-1)}=\frac{47}{60}
\end{align*}

Hoping for a simple proportion we equate the numerators and get
  \begin{align*}
3n^2-1=47\quad\Rightarrow\quad 3n^2=48\quad\Rightarrow\quad n\in\{4,-4\}
\end{align*}
The denominator $n(n^2-1)$ evaluated at $n=4$ gives $4(16-1)=60$ and we get the solution $$\color{blue}{n=4}$$

Note: Observe, it  was the symmetry which enabled us to make the lucky guess.
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