Question

Find the sum of all positive integers x such that
$\dfrac{{{x}^{3}}-x+120}{\left( x-1 \right)\left( x+1 \right)}$ is an integer.

Answer 1

Hint: To solve this question, we will first separate both terms $\dfrac{{{x}^{3}}-x}{\left( x-1 \right)\left( x+1 \right)}\text{ and }\dfrac{120}{\left( x-1 \right)\left( x+1 \right)}$ and then factorize the first term using identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$
Then, we will assume the left term as integer and see all values of x possible. Finally we will add all those obtained values of x which are positive integers.

Complete step-by-step answer:
We have $\mathbb{Z}$ denote set of integers and we need
\[\dfrac{{{x}^{3}}-x+120}{\left( x-1 \right)\left( x+1 \right)}\] to be an integer
\[\dfrac{{{x}^{3}}-x+120}{\left( x-1 \right)\left( x+1 \right)}\in \mathbb{Z}\] where x is also an integer.
Let us simplify the given term and take x common in the numerator.
\[\begin{align}
  & \dfrac{x\left( {{x}^{2}}-1 \right)+120}{\left( x-1 \right)\left( x+1 \right)}\in \mathbb{Z}\text{ where x}\in \mathbb{Z} \\
 & \dfrac{x\left( {{x}^{2}}-1 \right)}{\left( x-1 \right)\left( x+1 \right)}+\dfrac{120}{\left( x-1 \right)\left( x+1 \right)}\in \mathbb{Z} \\
\end{align}\]
Now, using the identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ in above we get:
\[\dfrac{x\left( x-1 \right)\left( x+1 \right)}{\left( x-1 \right)\left( x+1 \right)}+\dfrac{120}{\left( x-1 \right)\left( x+1 \right)}\in \mathbb{Z}\]
Cancelling the term $\left( x-1 \right)\left( x+1 \right)$ from above equation, we get:
\[x+\dfrac{120}{\left( x-1 \right)\left( x+1 \right)}\in \mathbb{Z}\]
Now, as x is also an integer and sum of two integers is also an integer.
For above term to be integer we should have \[\dfrac{120}{\left( x-1 \right)\left( x+1 \right)}\in \mathbb{Z}\]
Now, $\dfrac{120}{\left( x-1 \right)\left( x+1 \right)}$ is an integer if and only if the term $\left( x-1 \right)\left( x+1 \right)$ is a factor of 120, so that then there is no term left in denominator.
From above stated identity we have $\left( x-1 \right)\left( x+1 \right)={{x}^{2}}-1$
${{x}^{2}}-1$ is a factor of 120.
The factors of 120 are \[\to 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120.\]
${{x}^{2}}-1$ has possible values are \[\to 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120.\]
Calculating x separately for all of them we have:
Case I ${{x}^{2}}-1=1$
\[\begin{align}
  & {{x}^{2}}-1=1 \\
 & {{x}^{2}}=2 \\
 & x=\pm \sqrt{2} \\
\end{align}\]
But $\pm \sqrt{2}$ are not integers, so case I is not possible.
Case II ${{x}^{2}}-1=2$
\[\begin{align}
  & {{x}^{2}}=2+1 \\
 & {{x}^{2}}=3 \\
 & x=\pm \sqrt{3} \\
\end{align}\]
Again not integer, so not possible.
Case III ${{x}^{2}}-1=3$
\[\begin{align}
  & {{x}^{2}}=3+1 \\
 & {{x}^{2}}=4 \\
 & x=\pm 2 \\
\end{align}\]
Now $\pm 2$ both are integers but we only need positive integers.
\[x=2\text{ is valid }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
Case IV ${{x}^{2}}-1=4$
\[\begin{align}
  & {{x}^{2}}=4+1 \\
 & {{x}^{2}}=5 \\
 & x=\pm \sqrt{5} \\
\end{align}\]
Not an integer, so not possible.
Case V ${{x}^{2}}-1=5$
\[\begin{align}
  & {{x}^{2}}=5+1 \\
 & {{x}^{2}}=6 \\
 & x=\pm \sqrt{6} \\
\end{align}\]
It is not an integer, hence not possible.
Case VI ${{x}^{2}}-1=6$
\[\begin{align}
  & {{x}^{2}}=6+1 \\
 & {{x}^{2}}=7 \\
 & x=\pm \sqrt{7} \\
\end{align}\]
It is not an integer, so it is not possible.
Case VII ${{x}^{2}}-1=8$
\[\begin{align}
  & {{x}^{2}}=8+1 \\
 & {{x}^{2}}=9 \\
 & x=\pm 3 \\
\end{align}\]
All integers but we only need positive.
\[x=3\text{ is valid }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
Case VIII ${{x}^{2}}-1=10$
\[\begin{align}
  & {{x}^{2}}=10+1 \\
 & {{x}^{2}}=11 \\
 & x=\pm \sqrt{11} \\
\end{align}\]
It is not an integer, so it is not possible.
Case IX ${{x}^{2}}-1=12$
\[\begin{align}
  & {{x}^{2}}=12+1 \\
 & {{x}^{2}}=13 \\
 & x=\pm \sqrt{13} \\
\end{align}\]
Not an integer, so it is not possible.
Case X ${{x}^{2}}-1=15$
\[\begin{align}
  & {{x}^{2}}=15+1 \\
 & {{x}^{2}}=16 \\
 & x=\pm 4 \\
\end{align}\]
We only need positive integers.
\[x=4\text{ is valid }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}\]
Case XI ${{x}^{2}}-1=20$
\[\begin{align}
  & {{x}^{2}}=20+1 \\
 & {{x}^{2}}=21 \\
 & x=\pm \sqrt{21} \\
\end{align}\]
Not integer, hence not possible.
Case XII ${{x}^{2}}-1=24$
\[\begin{align}
  & {{x}^{2}}=24+1 \\
 & {{x}^{2}}=25 \\
 & x=\pm 5 \\
\end{align}\]
We only need positive integers.
\[x=5\text{ is valid }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iv)}\]
Case XIII ${{x}^{2}}-1=30$
\[\begin{align}
  & {{x}^{2}}=30+1 \\
 & {{x}^{2}}=31 \\
 & x=\pm \sqrt{31} \\
\end{align}\]
It is not an integer, hence it is not possible.
Case XIV ${{x}^{2}}-1=40$
\[\begin{align}
  & {{x}^{2}}=40+1 \\
 & {{x}^{2}}=41 \\
 & x=\pm \sqrt{41} \\
\end{align}\]
Again, we get that it is not an integer and it is not possible.
Case XV ${{x}^{2}}-1=60$
\[\begin{align}
  & {{x}^{2}}=60+1 \\
 & {{x}^{2}}=61 \\
 & x=\pm \sqrt{61} \\
\end{align}\]
Since it is not an integer, we cannot consider it.
Case XVI ${{x}^{2}}-1=120$
\[\begin{align}
  & {{x}^{2}}=120+1 \\
 & {{x}^{2}}=121 \\
 & x=\pm \sqrt{121} \\
\end{align}\]
Not an integer, so not possible.
So from equation (i), (ii), (iii) and (iv) we have all positive integers of x such that $\dfrac{{{x}^{3}}-x+120}{\left( x-1 \right)\left( x+1 \right)}$ an integer are 2, 3, 4, 5.
Sum of all above \[\Rightarrow 2+3+4+5=14\]
Therefore, the sum of all positive integer x such that $\dfrac{{{x}^{3}}-x+120}{\left( x-1 \right)\left( x+1 \right)}$ is an integer is 14.

Note: Students might get confused at the part where only $\dfrac{120}{\left( x-1 \right)\left( x+1 \right)}$ is considered to find all possible values of x and not the whole term $x+\dfrac{120}{\left( x-1 \right)\left( x+1 \right)}$
This is so because the sum and difference of integers is also integer.
For understanding let $x+\dfrac{120}{\left( x-1 \right)\left( x+1 \right)}=t$ where t is an integer, subtracting x from both sides of equation we get:
\[\dfrac{120}{\left( x-1 \right)\left( x+1 \right)}=t-x\]
Now, as t is integer and we need x also integer, so as the difference of integer is integer. Hence, we can directly go for calculating the value of x such that $\dfrac{120}{\left( x-1 \right)\left( x+1 \right)}$ is an integer.